📖Topic Explanations

🌐 Overview
Hello students! Welcome to Integration by Partial Fractions! Get ready to unlock a powerful technique that will transform seemingly daunting integration problems into manageable steps.

Imagine you're faced with an integral of a complex rational function – a fraction where both the numerator and denominator are polynomials. Traditional substitution or direct formulas might not work. It feels like trying to eat a giant cake in one bite! How do you simplify such expressions to make them integrable?

This is precisely where Partial Fractions comes to your rescue! It's a brilliant algebraic strategy that allows us to decompose a complex rational expression into a sum of simpler, more elementary rational expressions. Think of it as breaking down that giant cake into bite-sized pieces, each much easier to handle and, crucially, much easier to integrate individually.

Consider it like reverse engineering the process of combining fractions. When you add $frac{1}{x-1} + frac{2}{x+2}$, you get a single complex fraction. Partial fractions teaches us how to go back from that complex fraction to its original, simpler components. Each of these simpler components is then easily integrated using standard formulas, often involving logarithms.

In this topic, you'll learn how to identify different forms of rational functions – those with non-repeated linear factors, repeated linear factors, non-repeated irreducible quadratic factors, and repeated irreducible quadratic factors in the denominator. For each type, we'll discover a systematic way to express the original fraction as a sum of partial fractions, each having a much simpler denominator.

The mastery of partial fractions is absolutely essential for both your board examinations and the JEE Main & Advanced. It's not just a standalone topic; it often serves as a crucial intermediate step in solving more complex problems across calculus, differential equations, and even in certain applications of physics and engineering. Many integrals that appear complex at first glance become trivial once you apply this decomposition technique.

By understanding this method, you'll gain the ability to tackle a wide range of integration problems involving rational functions with confidence and precision. It’s a powerful tool that simplifies complexity, making your journey through calculus much smoother and more enjoyable. Get ready to add a fantastic weapon to your integration arsenal!

So, let's dive in and unravel the elegant mechanics of Integration by Partial Fractions, transforming challenging integrals into conquerable quests!
📚 Fundamentals
Hello, aspiring mathematicians! Welcome to another exciting session where we unravel the mysteries of Integral Calculus. Today, we're diving into a super important and incredibly useful technique called Integration by Partial Fractions.

Now, I know what you might be thinking: "Another integration technique? Haven't we learned enough?" Trust me, this one is a game-changer, especially when you encounter integrals of a specific type. Let's get started!

### What is Integration by Partial Fractions? The Big Picture

Imagine you're given a complex fraction, something like this:
$$ int frac{2x+1}{x^2+x-2} ,dx $$

Your first thought might be substitution, right? Let $u = x^2+x-2$, then $du = (2x+1)dx$. This works perfectly here! The integral becomes $int frac{1}{u} du = ln|u| + C = ln|x^2+x-2| + C$.
Great! But what if the numerator wasn't exactly the derivative of the denominator?

Consider this slightly different integral:
$$ int frac{x+5}{x^2+x-2} ,dx $$

Now, if you try $u = x^2+x-2$, $du = (2x+1)dx$. The numerator is $x+5$, which is not a direct multiple of $2x+1$. So, a simple substitution won't work here. This is where we need a different approach.

Think of it like this: Sometimes, a big, complex problem becomes much easier to solve if you can break it down into smaller, simpler, more manageable pieces. Partial fraction decomposition is exactly that for rational functions. It's a technique to break down a complex rational function (a fraction where both numerator and denominator are polynomials) into a sum of simpler fractions that are much easier to integrate individually.

The core idea is to reverse the process of adding fractions. Remember how you add $frac{1}{x-1} + frac{1}{x+2}$?
$$ frac{1}{x-1} + frac{1}{x+2} = frac{(x+2) + (x-1)}{(x-1)(x+2)} = frac{2x+1}{x^2+x-2} $$
Integration by partial fractions aims to do the opposite: if you start with $frac{2x+1}{x^2+x-2}$, you want to convert it back into $frac{1}{x-1} + frac{1}{x+2}$. Why? Because integrals of $frac{1}{x-1}$ and $frac{1}{x+2}$ are simply $ln|x-1|$ and $ln|x+2|$ respectively, which are super easy!

### Prerequisite: Understanding Rational Functions

Before we jump into the "how-to," let's quickly recap what a rational function is, as this technique applies specifically to them.

A rational function is simply a function that can be expressed as the ratio of two polynomials, say $P(x)$ and $Q(x)$, where $Q(x)$ is not the zero polynomial. We write it as $frac{P(x)}{Q(x)}$.

For example:
* $frac{x+5}{x^2+x-2}$ is a rational function. $P(x) = x+5$, $Q(x) = x^2+x-2$.
* $frac{1}{x^2-9}$ is a rational function.
* $frac{x^3+2x}{x^2+1}$ is also a rational function.

Now, rational functions can be classified into two types:

1. Proper Rational Functions: A rational function $frac{P(x)}{Q(x)}$ is called proper if the degree of the numerator polynomial $P(x)$ is strictly less than the degree of the denominator polynomial $Q(x)$.
* Example: $frac{x+5}{x^2+x-2}$ (degree of numerator is 1, degree of denominator is 2; 1 < 2).
* These are the functions we directly apply partial fractions to.

2. Improper Rational Functions: A rational function $frac{P(x)}{Q(x)}$ is called improper if the degree of the numerator polynomial $P(x)$ is greater than or equal to the degree of the denominator polynomial $Q(x)$.
* Example: $frac{x^3+2x}{x^2+1}$ (degree of numerator is 3, degree of denominator is 2; 3 >= 2).
* Important Note: You cannot apply partial fraction decomposition directly to an improper rational function. First, you must perform polynomial long division to convert it into a sum of a polynomial and a proper rational function.

Let's quickly see how that works with an example: $frac{x^3+2x}{x^2+1}$
```
x
_______
x^2+1 | x^3 + 2x
-(x^3 + x)
_________
x
```
So, $frac{x^3+2x}{x^2+1} = x + frac{x}{x^2+1}$.
Now, $frac{x}{x^2+1}$ is a proper rational function, and we can integrate $x$ easily.
We will mostly focus on proper rational functions in this section, as they are the direct beneficiaries of partial fraction decomposition.

### The Mechanism: Types of Denominators and Their Decomposition

The way we decompose a rational function into partial fractions depends entirely on the nature of the factors in its denominator, $Q(x)$. So, the first crucial step is always to factorize the denominator completely.

Let's look at the main cases:

#### Case 1: Denominator with Non-Repeated Linear Factors

If the denominator $Q(x)$ can be factored into distinct (non-repeated) linear factors like $(ax+b)(cx+d)...$, then the proper rational function $frac{P(x)}{Q(x)}$ can be expressed as a sum of partial fractions, each with a constant numerator over a linear factor.

For example, if $Q(x) = (x-a)(x-b)(x-c)$, then:
$$ frac{P(x)}{(x-a)(x-b)(x-c)} = frac{A}{x-a} + frac{B}{x-b} + frac{C}{x-c} $$
where $A, B, C$ are constants that we need to determine.

How to find A, B, C?
There are two main methods:

1. Equating Coefficients:
* Multiply both sides by the common denominator $Q(x)$ to get rid of the fractions.
* Expand the right side.
* Equate the coefficients of like powers of $x$ from both sides.
* Solve the resulting system of linear equations for $A, B, C$.

2. Substitution Method (Heaviside's Cover-Up Method): This is often quicker for non-repeated linear factors.
* Multiply both sides by the common denominator $Q(x)$.
* Substitute the roots of each linear factor into the equation. For example, to find $A$, substitute $x=a$ (the root of $x-a$). Since $(x-b)$ and $(x-c)$ terms will become zero, only $A$ will remain.

Let's try an example:
Example 1: Find $int frac{x+5}{x^2+x-2} ,dx$

Step 1: Check if proper. Yes, degree of numerator (1) < degree of denominator (2).
Step 2: Factorize the denominator.
$x^2+x-2 = (x+2)(x-1)$
Step 3: Write the partial fraction decomposition.
$$ frac{x+5}{(x+2)(x-1)} = frac{A}{x+2} + frac{B}{x-1} $$
Step 4: Find A and B.
Multiply by $(x+2)(x-1)$:
$$ x+5 = A(x-1) + B(x+2) quad ext{(Equation 1)} $$
Using Substitution Method:
* To find $A$, let $x = -2$ (the root of $x+2$):
$-2+5 = A(-2-1) + B(-2+2)$
$3 = A(-3) + 0 implies -3A = 3 implies mathbf{A = -1}$
* To find $B$, let $x = 1$ (the root of $x-1$):
$1+5 = A(1-1) + B(1+2)$
$6 = 0 + B(3) implies 3B = 6 implies mathbf{B = 2}$

So, our decomposition is:
$$ frac{x+5}{x^2+x-2} = frac{-1}{x+2} + frac{2}{x-1} $$
Step 5: Integrate each simpler fraction.
$$ int frac{x+5}{x^2+x-2} ,dx = int left( frac{-1}{x+2} + frac{2}{x-1}
ight) ,dx $$
$$ = - int frac{1}{x+2} ,dx + 2 int frac{1}{x-1} ,dx $$
$$ = -ln|x+2| + 2ln|x-1| + C $$
Using logarithm properties, this can also be written as $lnleft|frac{(x-1)^2}{x+2}
ight| + C$.
See how much easier that was?

#### Case 2: Denominator with Repeated Linear Factors

If the denominator $Q(x)$ has a repeated linear factor, say $(ax+b)^n$, then for each such factor, the decomposition must include $n$ partial fractions with increasing powers of that factor in the denominator, up to $n$.

For example, if $Q(x) = (x-a)^3(x-b)$, then:
$$ frac{P(x)}{(x-a)^3(x-b)} = frac{A}{x-a} + frac{B}{(x-a)^2} + frac{C}{(x-a)^3} + frac{D}{x-b} $$

How to find A, B, C, D?
The substitution method works partially. You can find $C$ by substituting $x=a$ and $D$ by substituting $x=b$. For $A$ and $B$, you'll typically use the equating coefficients method after finding $C$ and $D$, or substitute other convenient values of $x$.

Example 2: Find $int frac{x}{(x-1)^2(x+1)} ,dx$

Step 1: Check if proper. Yes, degree of numerator (1) < degree of denominator (3).
Step 2: Denominator is already factored.
Step 3: Write the partial fraction decomposition.
$$ frac{x}{(x-1)^2(x+1)} = frac{A}{x-1} + frac{B}{(x-1)^2} + frac{C}{x+1} $$
Step 4: Find A, B, and C.
Multiply by $(x-1)^2(x+1)$:
$$ x = A(x-1)(x+1) + B(x+1) + C(x-1)^2 quad ext{(Equation 2)} $$
Using Substitution Method (partially):
* To find $B$, let $x = 1$ (root of $(x-1)$):
$1 = A(0)(2) + B(1+1) + C(0)^2$
$1 = 2B implies mathbf{B = 1/2}$
* To find $C$, let $x = -1$ (root of $(x+1)$):
$-1 = A(-1-1)(-1+1) + B(-1+1) + C(-1-1)^2$
$-1 = A(0) + B(0) + C(-2)^2$
$-1 = 4C implies mathbf{C = -1/4}$

Now we have $B=1/2$ and $C=-1/4$. We need $A$. We can use equating coefficients or substitute another value for $x$.
Let's use equating coefficients for $A$. Expand Equation 2:
$$ x = A(x^2-1) + B(x+1) + C(x^2-2x+1) $$
$$ x = Ax^2 - A + Bx + B + Cx^2 - 2Cx + C $$
Group terms by powers of $x$:
$$ x = (A+C)x^2 + (B-2C)x + (-A+B+C) $$
Equate coefficients of $x^2$:
Left side: $0x^2$, Right side: $(A+C)x^2$
$0 = A+C implies A = -C$
Since $C = -1/4$, then $mathbf{A = 1/4}$.
(You could also equate coefficients of $x^0$ (constant term) or $x^1$ to verify or find if other unknowns were left.)

So, our decomposition is:
$$ frac{x}{(x-1)^2(x+1)} = frac{1/4}{x-1} + frac{1/2}{(x-1)^2} + frac{-1/4}{x+1} $$
Step 5: Integrate each simpler fraction.
$$ int frac{x}{(x-1)^2(x+1)} ,dx = int left( frac{1/4}{x-1} + frac{1/2}{(x-1)^2} - frac{1/4}{x+1}
ight) ,dx $$
$$ = frac{1}{4} int frac{1}{x-1} ,dx + frac{1}{2} int (x-1)^{-2} ,dx - frac{1}{4} int frac{1}{x+1} ,dx $$
$$ = frac{1}{4}ln|x-1| + frac{1}{2} frac{(x-1)^{-1}}{-1} - frac{1}{4}ln|x+1| + C $$
$$ = frac{1}{4}ln|x-1| - frac{1}{2(x-1)} - frac{1}{4}ln|x+1| + C $$
$$ = frac{1}{4}lnleft|frac{x-1}{x+1}
ight| - frac{1}{2(x-1)} + C $$

#### Case 3: Denominator with Non-Repeated Irreducible Quadratic Factors

If the denominator $Q(x)$ contains a non-repeated quadratic factor $ax^2+bx+c$ that cannot be factored into linear factors with real coefficients (i.e., its discriminant $b^2-4ac < 0$), then the corresponding partial fraction will have a linear numerator.

For example, if $Q(x) = (ax^2+bx+c)(dx+e)$, then:
$$ frac{P(x)}{(ax^2+bx+c)(dx+e)} = frac{Ax+B}{ax^2+bx+c} + frac{C}{dx+e} $$
Here, $A, B, C$ are constants.

How to find A, B, C?
You will generally rely on the equating coefficients method here. Substituting roots works for linear factors, but quadratic factors usually don't have real roots to substitute. You can find the constants for linear factors by substitution, then use equating coefficients for the rest.

Example 3: Find $int frac{x^2+1}{(x-1)(x^2+x+1)} ,dx$

Step 1: Check if proper. Yes, degree of numerator (2) < degree of denominator (3).
Step 2: Denominator is already factored. $x^2+x+1$ is irreducible because $b^2-4ac = 1^2 - 4(1)(1) = 1-4 = -3 < 0$.
Step 3: Write the partial fraction decomposition.
$$ frac{x^2+1}{(x-1)(x^2+x+1)} = frac{A}{x-1} + frac{Bx+C}{x^2+x+1} $$
Step 4: Find A, B, and C.
Multiply by $(x-1)(x^2+x+1)$:
$$ x^2+1 = A(x^2+x+1) + (Bx+C)(x-1) quad ext{(Equation 3)} $$
Using Substitution Method (partially):
* To find $A$, let $x = 1$ (root of $x-1$):
$1^2+1 = A(1^2+1+1) + (B(1)+C)(1-1)$
$2 = A(3) + 0 implies mathbf{A = 2/3}$

Now we have $A=2/3$. We need $B$ and $C$. Use equating coefficients:
Expand Equation 3:
$$ x^2+1 = Ax^2+Ax+A + Bx^2 - Bx + Cx - C $$
Group terms by powers of $x$:
$$ x^2+1 = (A+B)x^2 + (A-B+C)x + (A-C) $$
Equate coefficients of $x^2$:
$1 = A+B$
Since $A=2/3$, then $1 = 2/3 + B implies mathbf{B = 1/3}$.

Equate constant terms ($x^0$):
$1 = A-C$
Since $A=2/3$, then $1 = 2/3 - C implies C = 2/3 - 1 implies mathbf{C = -1/3}$.

(You can check with the coefficient of $x$: $0 = A-B+C implies 0 = 2/3 - 1/3 - 1/3 = 0$. It matches!)

So, our decomposition is:
$$ frac{x^2+1}{(x-1)(x^2+x+1)} = frac{2/3}{x-1} + frac{(1/3)x - 1/3}{x^2+x+1} $$
$$ = frac{2}{3(x-1)} + frac{x-1}{3(x^2+x+1)} $$
Step 5: Integrate each simpler fraction.
$$ int frac{x^2+1}{(x-1)(x^2+x+1)} ,dx = int frac{2}{3(x-1)} ,dx + int frac{x-1}{3(x^2+x+1)} ,dx $$
The first part is easy: $frac{2}{3}ln|x-1|$.
For the second part, $int frac{x-1}{3(x^2+x+1)} ,dx$, we often need to complete the square in the denominator and manipulate the numerator to match the derivative of the denominator.
Let $I_2 = frac{1}{3} int frac{x-1}{x^2+x+1} ,dx$.
The derivative of $x^2+x+1$ is $2x+1$. We need to manipulate $x-1$ to get $2x+1$.
$x-1 = frac{1}{2}(2x-2) = frac{1}{2}(2x+1 - 3)$
So, $I_2 = frac{1}{3} int frac{frac{1}{2}(2x+1-3)}{x^2+x+1} ,dx = frac{1}{6} int frac{2x+1}{x^2+x+1} ,dx - frac{3}{6} int frac{1}{x^2+x+1} ,dx$
$I_2 = frac{1}{6} ln|x^2+x+1| - frac{1}{2} int frac{1}{x^2+x+1} ,dx$.
For $int frac{1}{x^2+x+1} ,dx$, complete the square: $x^2+x+1 = (x^2+x+frac{1}{4}) + 1 - frac{1}{4} = (x+frac{1}{2})^2 + frac{3}{4}$.
So, $int frac{1}{(x+frac{1}{2})^2 + (frac{sqrt{3}}{2})^2} ,dx = frac{1}{frac{sqrt{3}}{2}} arctanleft(frac{x+frac{1}{2}}{frac{sqrt{3}}{2}}
ight) + C'$
$= frac{2}{sqrt{3}} arctanleft(frac{2x+1}{sqrt{3}}
ight) + C'$.
Putting it all together:
$$ int frac{x^2+1}{(x-1)(x^2+x+1)} ,dx = frac{2}{3}ln|x-1| + frac{1}{6} ln|x^2+x+1| - frac{1}{2} left( frac{2}{sqrt{3}} arctanleft(frac{2x+1}{sqrt{3}}
ight)
ight) + C $$
$$ = frac{2}{3}ln|x-1| + frac{1}{6} ln|x^2+x+1| - frac{1}{sqrt{3}} arctanleft(frac{2x+1}{sqrt{3}}
ight) + C $$
As you can see, integrating the quadratic factor part can be more involved, often leading to a mix of logarithms and inverse trigonometric functions.

#### Case 4: Denominator with Repeated Irreducible Quadratic Factors
This case is less common for basic problems but good to know for a complete understanding. If $Q(x)$ has a factor like $(ax^2+bx+c)^n$, then for each such factor, you'd include:
$$ frac{A_1x+B_1}{ax^2+bx+c} + frac{A_2x+B_2}{(ax^2+bx+c)^2} + dots + frac{A_nx+B_n}{(ax^2+bx+c)^n} $$
Integrals arising from $(ax^2+bx+c)^n$ for $n>1$ can be complex and often require reduction formulas or specific substitutions, which are usually covered in advanced topics. For fundamentals, understanding the setup is key.

### Summary of Steps for Integration by Partial Fractions:

1. Check if Proper: Ensure the rational function $frac{P(x)}{Q(x)}$ is proper (degree of $P(x)$ < degree of $Q(x)$). If not, perform polynomial long division first.
2. Factorize Denominator: Completely factorize the denominator $Q(x)$ into linear and/or irreducible quadratic factors.
3. Decompose: Set up the partial fraction decomposition based on the types of factors in $Q(x)$:
* For each distinct linear factor $(ax+b)$: include $frac{A}{ax+b}$.
* For each repeated linear factor $(ax+b)^n$: include $frac{A_1}{ax+b} + frac{A_2}{(ax+b)^2} + dots + frac{A_n}{(ax+b)^n}$.
* For each distinct irreducible quadratic factor $(ax^2+bx+c)$: include $frac{Ax+B}{ax^2+bx+c}$.
* For each repeated irreducible quadratic factor $(ax^2+bx+c)^n$: include $frac{A_1x+B_1}{ax^2+bx+c} + dots + frac{A_nx+B_n}{(ax^2+bx+c)^n}$.
4. Find Constants: Solve for the unknown constants (A, B, C, etc.) by clearing denominators and using either the substitution method (for linear factors) or equating coefficients of like powers of $x$.
5. Integrate: Integrate each of the simpler partial fractions obtained. Remember that linear factors lead to $ln|...|$ terms, and irreducible quadratic factors often lead to $ln|...|$ and $arctan(...)$ terms.

### CBSE vs. JEE Focus: What to Expect































Aspect CBSE (Class XII) JEE (Mains & Advanced)
Types of Factors Primarily focuses on non-repeated and repeated linear factors. Simple non-repeated irreducible quadratic factors (like $x^2+a^2$) are also covered. All types of factors are expected, including complex combinations, higher powers of repeated factors, and more general irreducible quadratic factors ($ax^2+bx+c$).
Improper Rational Functions Typically, only simple polynomial long division is required before decomposition. Expect problems where polynomial long division might be required, sometimes with higher degree numerators/denominators.
Integration Complexity Integrals usually simplify to standard forms like $int frac{1}{ax+b} dx$ or $int frac{1}{x^2+a^2} dx$. Integration of decomposed fractions can be more involved, often requiring completing the square, substitutions (e.g., $t= an(x/2)$ for trigonometric integrals reducible to rational functions), or other advanced techniques in conjunction with partial fractions.
Problem-Solving Steps Straightforward application of the method. Focus is on accurate decomposition and integration. Often requires a clever substitution *before* applying partial fractions (e.g., $e^x=t$, $sin x = t$, etc.) to convert a non-rational integral into a rational one, then applying partial fractions. This adds an extra layer of complexity.


For JEE, a strong grasp of the fundamentals, especially factorization and solving for constants, is crucial. But you also need to be adept at recognizing when a substitution can transform an integral into a form suitable for partial fractions, and then handling the subsequent, possibly complex, integrations.

That's a thorough introduction to the fundamentals of Integration by Partial Fractions! Mastering this technique opens up a whole new world of integrals you can solve. Practice these cases, and you'll be well on your way to conquering more complex problems. Keep up the great work!
🔬 Deep Dive
Welcome back, future IITians! In our journey through Integral Calculus, we've encountered various techniques like direct integration, substitution, and integration by parts. Today, we're diving deep into another incredibly powerful and indispensable technique: Integration by Partial Fractions. This method is your go-to solution for integrating complex rational functions, which are often challenging or impossible to integrate directly.

Understanding the 'Why': The Need for Partial Fractions


Recall a rational function is a function that can be expressed as the ratio of two polynomials, say $P(x)/Q(x)$, where $Q(x)
eq 0$.
For example, $frac{1}{x^2-1}$ or $frac{3x+2}{x^3+x}$.

You can easily integrate $frac{1}{x}$ (which is $ln|x|+C$) or $frac{1}{x^2+1}$ (which is $ an^{-1}(x)+C$). But what about something like $int frac{1}{x^2-1} dx$? It doesn't immediately fit a standard form.
This is where partial fractions come in! The core idea is to break down a complicated rational function into a sum of simpler rational functions, each of which we *can* integrate using standard formulas. Think of it like dismantling a complex machine into its basic components so you can fix or understand each part individually.

Before we proceed, let's briefly recall two crucial prerequisites:

  1. Proper vs. Improper Rational Functions:

    • A rational function $P(x)/Q(x)$ is a proper rational function if the degree of the numerator $P(x)$ is strictly less than the degree of the denominator $Q(x)$. (e.g., $frac{x}{x^2+1}$)

    • It is an improper rational function if the degree of $P(x)$ is greater than or equal to the degree of $Q(x)$. (e.g., $frac{x^3+1}{x^2+1}$)


    JEE Focus: Always remember that partial fraction decomposition can only be directly applied to proper rational functions. If you encounter an improper one, your first step *must* be polynomial long division!

  2. Polynomial Factorization: The success of partial fractions heavily relies on your ability to factorize the denominator $Q(x)$ completely.



The Core Idea: Partial Fraction Decomposition


The process involves expressing a given rational function $P(x)/Q(x)$ as a sum of simpler fractions whose denominators are the factors of $Q(x)$. The specific form of these simpler fractions depends on the nature of the factors in the denominator $Q(x)$.

Let's explore the various cases for the factors of the denominator $Q(x)$:

Case 1: Denominator has Distinct Linear Factors


If $Q(x)$ can be factored into distinct linear factors, say $(ax+b)(cx+d)(ex+f)$, then the rational function can be expressed as:
$$ frac{P(x)}{(ax+b)(cx+d)(ex+f)} = frac{A}{ax+b} + frac{B}{cx+d} + frac{C}{ex+f} $$
where $A, B, C$ are constants that we need to determine.

Method to Find Constants:
There are two primary ways:

  1. Comparing Coefficients: Multiply both sides by the common denominator $Q(x)$ to clear the fractions. Expand the right side and collect terms based on powers of $x$. Equate the coefficients of corresponding powers of $x$ on both sides to form a system of linear equations, then solve for $A, B, C$.

  2. Substituting Specific Values of $x$: After clearing fractions, choose values of $x$ that make each linear factor zero (e.g., $x = -b/a$, $x = -d/c$, etc.). This often makes terms disappear, directly yielding the values of $A, B, C$. This is often called the Heaviside Cover-up Method for linear factors.


Tip: A combination of both methods is often the most efficient for JEE problems.

Example 1: Integrate $int frac{x+5}{x^2+x-2} dx$.
First, factor the denominator: $x^2+x-2 = (x+2)(x-1)$.
So, we set up the partial fraction decomposition:
$$ frac{x+5}{(x+2)(x-1)} = frac{A}{x+2} + frac{B}{x-1} $$
Multiply both sides by $(x+2)(x-1)$:
$$ x+5 = A(x-1) + B(x+2) quad (*)$$

To find $A$ and $B$:
Substitute $x=1$ (makes $x-1$ zero):
$1+5 = A(1-1) + B(1+2)$
$6 = 0 + 3B implies B=2$

Substitute $x=-2$ (makes $x+2$ zero):
$-2+5 = A(-2-1) + B(-2+2)$
$3 = -3A + 0 implies A=-1$

So, the decomposition is:
$$ frac{x+5}{(x+2)(x-1)} = frac{-1}{x+2} + frac{2}{x-1} $$
Now, integrate:
$$ int left( frac{-1}{x+2} + frac{2}{x-1}
ight) dx = -int frac{1}{x+2} dx + 2int frac{1}{x-1} dx $$
$$ = -ln|x+2| + 2ln|x-1| + C $$
$$ = lnleft|frac{(x-1)^2}{x+2}
ight| + C $$

Case 2: Denominator has Repeated Linear Factors


If $Q(x)$ has a linear factor $(ax+b)$ repeated $n$ times, i.e., $(ax+b)^n$, then for this factor, the partial fraction decomposition will include $n$ terms:
$$ frac{P(x)}{(ax+b)^n cdot ( ext{other factors})} = frac{A_1}{ax+b} + frac{A_2}{(ax+b)^2} + dots + frac{A_n}{(ax+b)^n} + dots $$
For example, if the denominator is $(x-1)^3(x+2)$, the form is:
$$ frac{A}{x-1} + frac{B}{(x-1)^2} + frac{C}{(x-1)^3} + frac{D}{x+2} $$

Example 2: Integrate $int frac{3x+1}{(x-1)^2(x+2)} dx$.
Set up the decomposition:
$$ frac{3x+1}{(x-1)^2(x+2)} = frac{A}{x-1} + frac{B}{(x-1)^2} + frac{C}{x+2} $$
Multiply by $(x-1)^2(x+2)$:
$$ 3x+1 = A(x-1)(x+2) + B(x+2) + C(x-1)^2 quad (**) $$

Substitute $x=1$:
$3(1)+1 = A(0) + B(1+2) + C(0)$
$4 = 3B implies B = 4/3$

Substitute $x=-2$:
$3(-2)+1 = A(0) + B(0) + C(-2-1)^2$
$-5 = C(-3)^2 = 9C implies C = -5/9$

To find $A$, we can either substitute another value for $x$ (like $x=0$) or compare coefficients. Let's compare the coefficient of $x^2$:
From $(**)$: $0x^2 + 3x + 1 = A(x^2+x-2) + B(x+2) + C(x^2-2x+1)$
Coefficient of $x^2$: $0 = A + C$.
Since $C = -5/9$, we have $A = -C = 5/9$.

So, the decomposition is:
$$ frac{3x+1}{(x-1)^2(x+2)} = frac{5/9}{x-1} + frac{4/3}{(x-1)^2} + frac{-5/9}{x+2} $$
Now, integrate:
$$ int left( frac{5}{9(x-1)} + frac{4}{3(x-1)^2} - frac{5}{9(x+2)}
ight) dx $$
$$ = frac{5}{9} ln|x-1| + frac{4}{3} int (x-1)^{-2} dx - frac{5}{9} ln|x+2| + C $$
$$ = frac{5}{9} ln|x-1| + frac{4}{3} frac{(x-1)^{-1}}{-1} - frac{5}{9} ln|x+2| + C $$
$$ = frac{5}{9} lnleft|frac{x-1}{x+2}
ight| - frac{4}{3(x-1)} + C $$

Case 3: Denominator has Non-Reducible Quadratic Factors


If $Q(x)$ has a quadratic factor $(ax^2+bx+c)$ that cannot be factored into linear factors with real coefficients (i.e., its discriminant $b^2-4ac < 0$), then the partial fraction corresponding to this factor will be of the form:
$$ frac{P(x)}{(ax^2+bx+c) cdot ( ext{other factors})} = frac{Ax+B}{ax^2+bx+c} + dots $$
Note: The numerator is a linear polynomial, not just a constant!

Example 3: Integrate $int frac{x^2+x+1}{(x-1)(x^2+1)} dx$.
Here, $x^2+1$ is an irreducible quadratic factor ($b^2-4ac = 0^2 - 4(1)(1) = -4 < 0$).
Set up the decomposition:
$$ frac{x^2+x+1}{(x-1)(x^2+1)} = frac{A}{x-1} + frac{Bx+C}{x^2+1} $$
Multiply by $(x-1)(x^2+1)$:
$$ x^2+x+1 = A(x^2+1) + (Bx+C)(x-1) quad (***) $$

Substitute $x=1$:
$1^2+1+1 = A(1^2+1) + (B(1)+C)(1-1)$
$3 = 2A + 0 implies A = 3/2$

Now, substitute $A=3/2$ into $(***)$ and equate coefficients.
$x^2+x+1 = frac{3}{2}(x^2+1) + Bx^2 - Bx + Cx - C$
$x^2+x+1 = left(frac{3}{2}+B
ight)x^2 + (C-B)x + left(frac{3}{2}-C
ight)$

Comparing coefficients of $x^2$:
$1 = frac{3}{2} + B implies B = 1 - frac{3}{2} = -frac{1}{2}$

Comparing coefficients of $x$:
$1 = C-B implies 1 = C - (-frac{1}{2}) implies 1 = C + frac{1}{2} implies C = frac{1}{2}$

(Check with constant term: $1 = frac{3}{2}-C implies 1 = frac{3}{2}-frac{1}{2} = 1$. It matches!)

So, the decomposition is:
$$ frac{x^2+x+1}{(x-1)(x^2+1)} = frac{3/2}{x-1} + frac{-1/2 x + 1/2}{x^2+1} = frac{3}{2(x-1)} + frac{1-x}{2(x^2+1)} $$
Now, integrate:
$$ int left( frac{3}{2(x-1)} + frac{1}{2(x^2+1)} - frac{x}{2(x^2+1)}
ight) dx $$
$$ = frac{3}{2} int frac{1}{x-1} dx + frac{1}{2} int frac{1}{x^2+1} dx - frac{1}{2} int frac{x}{x^2+1} dx $$
For the last integral, let $u=x^2+1 implies du=2x dx implies x dx = frac{1}{2} du$.
$$ = frac{3}{2}ln|x-1| + frac{1}{2} an^{-1}(x) - frac{1}{2} int frac{1}{u} frac{1}{2} du $$
$$ = frac{3}{2}ln|x-1| + frac{1}{2} an^{-1}(x) - frac{1}{4}ln|u| + C $$
$$ = frac{3}{2}ln|x-1| + frac{1}{2} an^{-1}(x) - frac{1}{4}ln(x^2+1) + C $$
Note: We write $ln(x^2+1)$ instead of $ln|x^2+1|$ because $x^2+1$ is always positive.

Case 4: Denominator has Repeated Non-Reducible Quadratic Factors


If $Q(x)$ has a quadratic factor $(ax^2+bx+c)$ repeated $n$ times, i.e., $(ax^2+bx+c)^n$, then for this factor, the partial fraction decomposition will include $n$ terms:
$$ frac{P(x)}{(ax^2+bx+c)^n cdot ( ext{other factors})} = frac{A_1x+B_1}{ax^2+bx+c} + frac{A_2x+B_2}{(ax^2+bx+c)^2} + dots + frac{A_nx+B_n}{(ax^2+bx+c)^n} + dots $$
This case can become algebraically intensive and is less frequent in standard JEE Mains, but it's important for advanced problems.
For example, if the denominator is $(x^2+1)^2(x-1)$, the form is:
$$ frac{A}{x-1} + frac{Bx+C}{x^2+1} + frac{Dx+E}{(x^2+1)^2} $$
The method to find constants is primarily by comparing coefficients, as substituting values that make quadratic factors zero is not as straightforward in real numbers.

JEE Advanced Focus: Handling Improper Rational Functions


As mentioned, if the degree of $P(x)$ is greater than or equal to the degree of $Q(x)$, you *must* perform polynomial long division first.
$$ frac{P(x)}{Q(x)} = S(x) + frac{R(x)}{Q(x)} $$
where $S(x)$ is the quotient polynomial, and $R(x)$ is the remainder polynomial, with $ ext{deg}(R(x)) < ext{deg}(Q(x))$.
Then, you integrate $S(x)$ (which is usually simple) and apply partial fractions to the proper rational function $R(x)/Q(x)$.

Example 4: Integrate $int frac{x^3+x^2+x+1}{x^2-1} dx$.
Here, degree of numerator (3) $ge$ degree of denominator (2). So, it's an improper rational function.
Perform polynomial long division:
$$ (x^3+x^2+x+1) div (x^2-1) $$
$$ x(x^2-1) = x^3-x $$
$$ (x^3+x^2+x+1) - (x^3-x) = x^2+2x+1 $$
$$ 1(x^2-1) = x^2-1 $$
$$ (x^2+2x+1) - (x^2-1) = 2x+2 $$
So, $frac{x^3+x^2+x+1}{x^2-1} = x+1 + frac{2x+2}{x^2-1}$.
Now we integrate:
$$ int left( x+1 + frac{2x+2}{x^2-1}
ight) dx = int (x+1) dx + int frac{2x+2}{(x-1)(x+1)} dx $$
The first part is easy: $int (x+1) dx = frac{x^2}{2} + x$.
For the second part: $frac{2x+2}{(x-1)(x+1)} = frac{2(x+1)}{(x-1)(x+1)} = frac{2}{x-1}$ (for $x
eq -1$).
So, $int frac{2}{x-1} dx = 2ln|x-1|$.
Thus, the final integral is $frac{x^2}{2} + x + 2ln|x-1| + C$.
Important: Be careful with cancellations. Here, $(x+1)$ cancelled, simplifying the partial fraction step significantly. If it didn't cancel, we'd proceed with standard partial fraction decomposition for $frac{2x+2}{(x-1)(x+1)}$.

Summary of Partial Fraction Forms


Here's a quick reference table for the partial fraction forms:































Factor in Denominator $Q(x)$ Form of Partial Fraction(s) Example
Distinct Linear Factor $(ax+b)$ $frac{A}{ax+b}$ $frac{P(x)}{(x-1)(x+2)} = frac{A}{x-1} + frac{B}{x+2}$
Repeated Linear Factor $(ax+b)^n$ $frac{A_1}{ax+b} + frac{A_2}{(ax+b)^2} + dots + frac{A_n}{(ax+b)^n}$ $frac{P(x)}{(x-1)^3} = frac{A}{x-1} + frac{B}{(x-1)^2} + frac{C}{(x-1)^3}$
Distinct Non-Reducible Quadratic Factor $(ax^2+bx+c)$ $frac{Ax+B}{ax^2+bx+c}$ $frac{P(x)}{(x-1)(x^2+4)} = frac{A}{x-1} + frac{Bx+C}{x^2+4}$
Repeated Non-Reducible Quadratic Factor $(ax^2+bx+c)^n$ $frac{A_1x+B_1}{ax^2+bx+c} + dots + frac{A_nx+B_n}{(ax^2+bx+c)^n}$ $frac{P(x)}{(x^2+1)^2} = frac{Ax+B}{x^2+1} + frac{Cx+D}{(x^2+1)^2}$


Integration of the Partial Fractions


Once the decomposition is done, the integration step uses standard formulas:

  • $int frac{A}{ax+b} dx = frac{A}{a} ln|ax+b| + C$

  • $int frac{A}{(ax+b)^n} dx = frac{A}{a} frac{(ax+b)^{-n+1}}{-n+1} + C$ (for $n
    eq 1$)

  • $int frac{Ax+B}{ax^2+bx+c} dx$: This one often requires splitting into two parts:

    • $int frac{Ax}{ax^2+bx+c} dx$: Use substitution ($u = ax^2+bx+c$, $du = (2ax+b)dx$). You might need to adjust the numerator to match $du$.

    • $int frac{B}{ax^2+bx+c} dx$: Complete the square in the denominator to get a form like $int frac{1}{(x+k)^2+p^2} dx$, which integrates to $frac{1}{p} an^{-1}left(frac{x+k}{p}
      ight) + C$.





JEE Advanced Context: Common Pitfalls & Strategies



  1. Forgetting Long Division: This is the most common mistake. Always check the degrees of the numerator and denominator.

  2. Incorrect Partial Fraction Forms: Ensure you use the correct setup for repeated factors and irreducible quadratic factors. Forgetting the $Ax+B$ form for quadratics is another frequent error.

  3. Algebraic Errors: Finding the constants $A, B, C, dots$ can be algebraically intensive. Be meticulous in comparing coefficients and solving systems of equations. Using a combination of substitution (for roots of linear factors) and comparing coefficients (for higher powers or constant terms) can minimize effort.

  4. Substitutions Before Partial Fractions: Sometimes, a simple substitution can transform a non-rational integral into a rational one suitable for partial fractions.

    Example: $int frac{cos x}{(1+sin x)(2+sin x)} dx$. Let $t=sin x implies dt=cos x dx$. The integral becomes $int frac{dt}{(1+t)(2+t)}$, which is a standard partial fraction problem.



Integration by partial fractions is a cornerstone technique in integral calculus. Master it thoroughly, and a wide array of integration problems will become approachable. Practice is key, especially with varied examples covering all the cases discussed! Keep exploring, keep learning, and you'll ace these concepts for JEE.
🎯 Shortcuts
Partial fractions is a powerful technique to integrate rational functions, especially those that don't directly fit standard integral forms. Mastering the method involves recognizing the types and efficiently finding the coefficients. Here are some mnemonics and shortcuts to help you remember and apply the concepts quickly in exams.

1. The Golden Rule: Proper Fraction Check


This is the absolute first step and a frequent source of errors if overlooked.

  • "D N L D" Mnemonic: Degree of Numerator Less than Degree of Denominator.

  • Shortcut: Before even thinking about partial fractions, always check if the degree of the polynomial in the numerator (N(x)) is strictly less than the degree of the polynomial in the denominator (D(x)).

    • If `deg(N(x)) < deg(D(x))`, it's a proper fraction. Proceed with partial fractions.

    • If `deg(N(x)) ≥ deg(D(x))`, it's an improper fraction. You MUST perform polynomial long division first. The result will be `Quotient + Remainder/D(x)`, where `Remainder/D(x)` will be a proper fraction.



  • JEE & CBSE Tip: Ignoring this step is a common mistake that leads to incorrect solutions. Always start here!



2. The Denominator's Blueprint: Factorization


The success of partial fractions relies entirely on factoring the denominator.

  • "PF Needs Factors!" Mnemonic: Partial Fractions Needs Factored Denominators.

  • Shortcut: Always factorize the denominator `D(x)` completely into linear and irreducible quadratic factors before attempting to set up the partial fraction decomposition.



3. Types of Factors and Their Forms


Understanding the decomposition form is crucial.

  • "L-L-Q-Q" Mnemonic for Forms:

    • Linear (Distinct): `A/(ax+b)`

    • Linear (Repeated): `A/(ax+b) + B/(ax+b)² + ...`

    • Quadratic (Distinct, Irreducible): `(Ax+B)/(ax²+bx+c)`

    • Quadratic (Repeated, Irreducible): `(Ax+B)/(ax²+bx+c) + (Cx+D)/(ax²+bx+c)² + ...`



  • Visualizing Shortcut: Imagine `(ax+b)` is a single block. If it's `(ax+b)²`, it means one block `(ax+b)` and two blocks `(ax+b)²`. For quadratics `(ax²+bx+c)`, the numerator must be one degree less than the denominator, hence `(Ax+B)`.



4. Finding Coefficients: Smart Strategies


This is where speed and accuracy make a huge difference.

The "R.S.E." Strategy (General Approach for all cases):

  • Roots: Substitute the roots of the linear factors into the equation (after clearing denominators). This is the fastest method.

  • Substitution: Substitute convenient values for `x` (like `x=0`, `x=1`, `x=-1`, etc.) into the equation after clearing denominators. This works well when roots are not easily available (e.g., for quadratic factors) or after finding some coefficients.

  • Equating: Equate coefficients of like powers of `x` (e.g., `x²`, `x`, constant term) on both sides of the equation after clearing denominators. This is a robust method, often used as a fallback or to find remaining coefficients.



Specific Shortcuts:

  • 1. Heaviside's "Cover-Up" Method (for Distinct Linear Factors only):
    * Mnemonic: "Cover up and Substitute!"
    * Shortcut: To find the coefficient `A` for a factor `(ax+b)`, simply `cover up` the `(ax+b)` term in the original denominator `D(x)`, and then substitute `x = -b/a` (the root of the covered factor) into the remaining expression `N(x) / [D(x) / (ax+b)]`.
    * Example: For `(3x+5) / [(x-1)(x+2)] = A/(x-1) + B/(x+2)`
    * To find A: Cover `(x-1)`, substitute `x=1` into `(3x+5)/(x+2)` -> `(3*1+5)/(1+2) = 8/3`. So, `A=8/3`.
    * To find B: Cover `(x+2)`, substitute `x=-2` into `(3x+5)/(x-1)` -> `(3*(-2)+5)/(-2-1) = -1/-3 = 1/3`. So, `B=1/3`.
    * JEE Tip: This is a HUGE time-saver for objective questions! Practice it thoroughly.

  • 2. For Repeated Linear Factors (e.g., `(ax+b)²`):
    * Shortcut: Use the "Cover-Up" method to find the coefficient of the highest power term (e.g., `B` for `B/(ax+b)²`). Then, use `x=0` or another convenient value for the remaining coefficients.
    * Example: For `(x+1) / (x-2)² = A/(x-2) + B/(x-2)²`
    * To find B: Cover `(x-2)²`, substitute `x=2` into `(x+1)` -> `(2+1) = 3`. So, `B=3`.
    * To find A: Substitute `x=0` into the original decomposition: `(0+1)/(0-2)² = A/(0-2) + B/(0-2)²`. We know `B=3`, so `1/4 = A/(-2) + 3/4`. This gives `A/(-2) = 1/4 - 3/4 = -2/4 = -1/2`. Hence, `A=1`.



5. Final Integration Step: Common Forms


Remember the standard integral forms after decomposition.

  • `∫ 1/(ax+b) dx = (1/a) ln|ax+b| + C`

  • `∫ 1/(ax+b)² dx = -1/[a(ax+b)] + C` (using `∫ u⁻² du`)

  • `∫ (Ax+B)/(ax²+bx+c) dx` often involves completing the square in the denominator and then using `ln` or `tan⁻¹` forms.



Keep practicing these mnemonics and shortcuts to build speed and accuracy. Happy integrating!
💡 Quick Tips
Here are some quick tips to master Integration by Partial Fractions, crucial for both JEE Main and CBSE board exams:


  1. Check for Proper Rational Function:

    • Critical First Step: Always ensure the given rational function $frac{P(x)}{Q(x)}$ is proper, meaning the degree of the numerator $P(x)$ must be less than the degree of the denominator $Q(x)$.

    • If it's an improper rational function (degree of $P(x) ge$ degree of $Q(x)$), first perform polynomial long division to express it as a polynomial plus a proper rational function.

      Example: $frac{x^3+x}{x^2-1} = x + frac{2x}{x^2-1}$

    • CBSE vs JEE: This step is fundamental for both, but JEE might present more complex improper fractions.



  2. Identify the Denominator Factor Type Quickly:

    • The form of partial fraction decomposition depends entirely on the factors of the denominator $Q(x)$. Master these standard forms:

























      Factor Type in $Q(x)$ Form of Partial Fraction Example
      Distinct Linear
      $(ax+b)(cx+d)$      		 $frac{A}{ax+b} + frac{B}{cx+d}$ $frac{2x+3}{(x-1)(x+2)} = frac{A}{x-1} + frac{B}{x+2}$
      Repeated Linear
      $(ax+b)^n$      		 $frac{A_1}{ax+b} + frac{A_2}{(ax+b)^2} + dots + frac{A_n}{(ax+b)^n}$ $frac{x}{(x+1)^2} = frac{A}{x+1} + frac{B}{(x+1)^2}$
      Irreducible Quadratic
      $(ax^2+bx+c)$      		 $frac{Ax+B}{ax^2+bx+c}$ $frac{x^2+1}{(x-1)(x^2+4)} = frac{A}{x-1} + frac{Bx+C}{x^2+4}$




  3. Efficient Coefficient Determination:

    • Substitution Method (Heaviside's "Cover-Up" Rule for Distinct Linear Factors): This is a massive time-saver for JEE. For factors like $(x-a)$, cover $(x-a)$ in the original expression and substitute $x=a$ into the remaining part to find the corresponding numerator coefficient.

      Example: To find $A$ for $frac{x}{(x-1)(x+2)} = frac{A}{x-1} + frac{B}{x+2}$, set $x=1$ in $frac{x}{(x+2)}$, so $A = frac{1}{1+2} = frac{1}{3}$.

    • Comparing Coefficients: For repeated linear or irreducible quadratic factors, or when the substitution method isn't enough, equate coefficients of powers of $x$ on both sides after multiplying by the common denominator.

    • Combination Strategy: Often, the fastest approach is to use substitution for distinct linear factors and then compare coefficients or use other convenient substitutions for the remaining unknown coefficients.



  4. Recognize Standard Integral Forms Post-Decomposition:

    • After decomposition, you will typically get integrals of the forms:

      • $int frac{1}{ax+b} dx = frac{1}{a} ln|ax+b| + C$

      • $int frac{1}{(ax+b)^n} dx = frac{1}{a} frac{(ax+b)^{1-n}}{1-n} + C$ (for $n
        e 1$)

      • $int frac{Ax+B}{ax^2+bx+c} dx$: Often split into two parts: one involving $ln$ (from derivative of denominator) and another involving $ an^{-1}$ (by completing the square in the denominator). This is a common pattern for JEE.





  5. Don't Forget the Constant: Always add the constant of integration '+C' to your final answer. This is crucial for both board exams and competitive exams.


Keep practicing these methods, as speed and accuracy in partial fraction decomposition are key to successfully solving related integral problems in exams!

🧠 Intuitive Understanding

Welcome to the intuitive understanding of Integration by Partial Fractions! This technique is a cornerstone for integrating a specific class of functions, and grasping its essence will make the procedural steps much clearer.



The Core Idea: Breaking Down Complex Fractions


Imagine you have a complex fraction, say $frac{5}{6}$. You know this can be written as the sum of simpler fractions: $frac{1}{2} + frac{1}{3}$. Adding fractions is usually easier than trying to figure out what they came from. Partial fractions is essentially the reverse process of adding fractions.




  • The Problem: We often encounter integrals of rational functions, which are ratios of two polynomials, like $frac{P(x)}{Q(x)}$. When the denominator $Q(x)$ is a product of linear or quadratic factors, integrating $frac{P(x)}{Q(x)}$ directly can be very challenging or even impossible with standard integration formulas.


  • The Solution: Partial fractions allow us to decompose (break down) a complex rational function into a sum of simpler rational functions. These simpler functions have denominators that are the individual factors of the original denominator.



Why Does This Help for Integration?


The magic happens because these "simpler" fractions are much easier to integrate using elementary formulas:




  • Type 1: $frac{A}{ax+b}$

    Integrals of this form directly lead to logarithmic functions: $int frac{A}{ax+b} dx = frac{A}{a} ln|ax+b| + C$. This is a very common outcome of partial fraction decomposition.




  • Type 2: $frac{Bx+C}{ax^2+bx+c}$ (where $ax^2+bx+c$ is irreducible)

    Integrals of this form can be handled by completing the square in the denominator, leading to combinations of logarithmic and inverse tangent functions (e.g., $arctan(dots)$).




Therefore, instead of trying to integrate a single complicated fraction, we break it into a sum of terms, each of which we already know how to integrate.



Analogy: Deconstructing a Recipe


Think of the complex rational function as a final dish with many ingredients blended together. Trying to understand each ingredient's contribution from the final dish is hard. Partial fractions is like having a magical tool that separates the dish back into its original, identifiable ingredients. Once separated, you can easily tell what each ingredient is and how to 'process' it individually.



JEE & CBSE Relevance: This technique is absolutely crucial for both JEE Main/Advanced and CBSE board exams. Many problems involving integrals of rational functions will require partial fraction decomposition as the primary step. Mastering the intuitive understanding and the procedural aspects will unlock a significant portion of integral calculus problems.



In essence, partial fractions is a powerful algebraic trick that transforms an unmanageable integral into a sum of manageable ones. It's about simplifying the integrand to fit into standard integration patterns.

🌍 Real World Applications

Real-World Applications of Integration by Partial Fractions


While integration by partial fractions might seem like a purely theoretical mathematical technique, it serves as a powerful tool in various scientific and engineering disciplines. Its primary utility lies in simplifying the integration of complex rational functions, which frequently arise when modeling real-world phenomena involving rates of change, accumulation, or system responses.




1. Engineering (Control Systems & Electrical Engineering)




  • System Analysis: In control systems and electrical engineering, the behavior of systems (e.g., circuits, mechanical systems) is often described using transfer functions. These functions are typically rational expressions in the Laplace domain (s-domain). To determine the system's time-domain response (how current, voltage, or position changes over time), we often need to perform an inverse Laplace transform, which involves integrating complex rational functions.


  • Partial Fractions Role: Partial fraction decomposition breaks down these intricate rational transfer functions into simpler, integrable terms (e.g., terms like $A/(s-a)$ or $(Bs+C)/((s-a)^2+b^2)$). This simplification makes the inverse Laplace transform much more manageable, allowing engineers to analyze stability, transient response, and steady-state behavior of systems.


  • JEE Relevance: While direct application in engineering problems isn't usually tested in JEE, understanding this context highlights the technique's practical importance.



2. Population Dynamics (Logistic Growth Model)




  • Modeling Limited Growth: Simple exponential growth models don't account for resource limitations. The logistic growth model provides a more realistic representation of population growth, where the growth rate slows down as the population approaches a carrying capacity. This model is often described by a differential equation: $frac{dP}{dt} = kP(1 - frac{P}{K})$, where $P$ is population, $t$ is time, $k$ is growth rate, and $K$ is carrying capacity.


  • Partial Fractions Role: To solve this differential equation and find $P(t)$, we separate variables: $frac{dP}{P(1 - P/K)} = k dt$. The left-hand side involves integrating a rational function of $P$. Partial fraction decomposition transforms $frac{1}{P(1 - P/K)}$ into simpler terms like $frac{A}{P} + frac{B}{1 - P/K}$, making the integration straightforward and leading to the characteristic 'S'-shaped logistic growth curve.



3. Chemical Reaction Kinetics




  • Rate Laws: In chemistry, the rate at which reactants are consumed and products are formed in a chemical reaction is described by rate laws. For certain complex reactions, particularly those involving multiple steps or reversible reactions, the rate equations can lead to differential equations involving rational functions.


  • Partial Fractions Role: When these differential equations are solved to find the concentration of reactants or products as a function of time, integration of rational expressions often becomes necessary. Partial fractions allow the chemist to decompose these expressions into simpler forms, facilitating the integration and enabling the prediction of reactant/product concentrations over time. This is crucial for designing and optimizing industrial chemical processes.




Understanding these applications reinforces that mathematical tools like partial fractions are not just academic exercises but fundamental for solving practical problems across diverse fields.


🔄 Common Analogies

Common Analogies for Integration by Partial Fractions


Understanding complex mathematical concepts often becomes easier when we relate them to everyday scenarios. Integration by Partial Fractions, though seemingly intricate, has several relatable analogies that can help solidify your grasp on the technique.



The Essence: Breaking Down Complexity


At its core, integration by partial fractions is about decomposing a complicated rational function into a sum of simpler fractions that are much easier to integrate. Think of it as a "reverse common denominator" process.



Analogy 1: Disassembling a LEGO Model



  • Imagine you have a large, complex LEGO model (your complicated rational function, e.g., $frac{5x-7}{x^2-2x-3}$).

  • It's difficult to analyze or modify individual pieces while they are part of the larger, assembled structure.

  • Partial fraction decomposition is like taking the LEGO model apart into its individual, simpler bricks (e.g., $frac{2}{x-3} - frac{3}{x+1}$).

  • Once you have the individual bricks, each one is very simple to handle or understand. Similarly, each partial fraction is easy to integrate using standard formulas like $int frac{1}{ax+b} dx = frac{1}{a} ln|ax+b| + C$.



Analogy 2: Unmixing a Smoothie



  • Consider a delicious smoothie made from various fruits (e.g., mango, banana, and yogurt). The final mixed smoothie represents your complex rational function.

  • If you were asked to integrate the "smoothie" directly, it would be a baffling task – how do you integrate a blend?

  • Partial fraction decomposition is akin to "unmixing" the smoothie back into its individual ingredients (mango pulp, banana puree, yogurt).

  • Each individual ingredient is a simple, distinct component that you know how to handle or measure separately. Mathematically, these are the simpler fractions, each of which has a known integral form.



Analogy 3: Reverse Engineering a Product



  • Suppose you have a sophisticated electronic gadget (your complex rational function) and you need to understand its internal workings or repair a specific part.

  • It's nearly impossible to do this from the outside or by treating it as a single unit.

  • Partial fraction decomposition is like reverse engineering the gadget: carefully taking it apart into its sub-assemblies and individual components (resistors, capacitors, wires).

  • Each component or sub-assembly (partial fraction) is much simpler to understand, test, and deal with individually. Integrating these simpler components is straightforward.



Why These Analogies Matter for Exams:



  • These analogies help you remember the fundamental purpose: to transform a difficult problem into a series of easier ones.

  • This strategic thinking is crucial for JEE, where problems often require choosing the most efficient method. If you see a rational function, mentally "disassemble" it into simpler parts.

  • While CBSE exams will primarily test your ability to execute the partial fraction decomposition steps for standard cases, JEE Main might present more complex denominators requiring careful factorization and a strong understanding of how to set up the partial fractions correctly (e.g., repeated factors, irreducible quadratic factors). The analogy reinforces why you are doing it.



Remember, sometimes the most complex problems are just combinations of simpler ones. Master the art of breaking them down!


📋 Prerequisites
To effectively tackle Integration by Partial Fractions, a strong foundation in specific algebraic and basic calculus concepts is essential. This section outlines the prerequisite knowledge you must possess to confidently apply this advanced integration technique.

Before diving into the integration aspect, ensure you are proficient in the following areas:




  • Algebraic Manipulation and Factorization:

    • Polynomial Factorization: You must be adept at factoring polynomial expressions in the denominator into linear and irreducible quadratic factors. This includes techniques like splitting the middle term, factor theorem, and recognizing common algebraic identities. This is a fundamental step for setting up partial fractions.

    • Solving Systems of Linear Equations: Once partial fractions are set up, you will need to solve for unknown constants (e.g., A, B, C). This typically involves solving systems of linear equations, either by substitution, elimination, or by equating coefficients of like powers of x.

    • Polynomial Division: For JEE Main & Advanced, it's crucial to understand that partial fraction decomposition is only directly applicable to proper rational functions (degree of numerator < degree of denominator). If the rational function is improper (degree of numerator ≥ degree of denominator), you must first perform polynomial long division to express it as a sum of a polynomial and a proper rational function.



  • Understanding Rational Functions:

    • Definition: Be clear on what constitutes a rational function, i.e., a function that can be expressed as the ratio of two polynomials, P(x)/Q(x).

    • Proper vs. Improper Rational Functions: Differentiate between proper rational functions (deg P(x) < deg Q(x)) and improper rational functions (deg P(x) ≥ deg Q(x)). This distinction dictates whether polynomial division is needed.



  • Basic Integration Formulas:

    • You must have a solid grasp of fundamental integration formulas, as the decomposed partial fractions will require their application. Key formulas include:

      • ∫ (1/x) dx = ln|x| + C

      • ∫ (1/(ax+b)) dx = (1/a) ln|ax+b| + C

      • ∫ (1/(x²+a²)) dx = (1/a) tan⁻¹(x/a) + C

      • ∫ ( (2x)/(x²+a²) ) dx = ln|x²+a²| + C (or generally, ∫ (f'(x)/f(x)) dx = ln|f(x)| + C)



    • Proficiency in these basic forms ensures that once the algebraic decomposition is complete, the integration step is straightforward.



  • Algebraic Partial Fraction Decomposition (without integration):

    • This is arguably the most critical prerequisite. You must already be proficient in the *algebraic technique* of decomposing a rational function into partial fractions based on the nature of the factors in the denominator:

      • Distinct Linear Factors: e.g., (A/(x-a)) + (B/(x-b))

      • Repeated Linear Factors: e.g., (A/(x-a)) + (B/(x-a)²) + (C/(x-a)³)

      • Irreducible Quadratic Factors: e.g., (Ax+B)/(ax²+bx+c)



    • The process of setting up the correct form and finding the constants A, B, C... is a prerequisite skill, forming the core of the method before actual integration begins.





Mastering these prerequisites will make the integration by partial fractions topic much more manageable and allow you to focus on the nuances of applying the integration formulas to the decomposed terms.

🔗 Related Topics

Related Topics to Integration by Partial Fractions



Integration by partial fractions is a powerful technique for integrating rational functions. To master this method and understand its broader context in calculus, it's essential to be familiar with several related topics. These concepts act as either prerequisites, foundational building blocks, or subsequent applications of partial fraction decomposition.

1. Algebra of Rational Functions: Polynomial Division & Factoring


This is a critical prerequisite.


  • Polynomial Division: Before applying partial fractions, ensure the rational function is a proper fraction (degree of numerator < degree of denominator). If it's an improper fraction (degree of numerator ≥ degree of denominator), you must perform polynomial long division first to express it as a sum of a polynomial and a proper rational function.

    JEE Tip: Skipping polynomial division for improper fractions is a common mistake that leads to incorrect partial fraction decomposition. Always check the degrees first.


  • Factoring Polynomials: The denominator of the proper rational function must be factored completely into linear and irreducible quadratic factors. This is the foundation for determining the form of the partial fraction decomposition. Proficiency in factoring quadratic expressions, using the rational root theorem, or synthetic division is crucial.



2. Basic Integration Formulas


The entire purpose of partial fraction decomposition is to transform a complex rational function into a sum of simpler fractions, each of which can be integrated using basic formulas.


  • Logarithmic Integral: Integrals of the form $int frac{1}{ax+b} dx = frac{1}{a} ln|ax+b| + C$ are the most common outcome from linear factors.


  • Inverse Tangent Integral: Integrals of the form $int frac{1}{x^2+a^2} dx = frac{1}{a} an^{-1}left(frac{x}{a}
    ight) + C$     arise from irreducible quadratic factors.


  • Power Rule and completing the square: Sometimes, terms like $frac{2x+A}{x^2+bx+c}$ might require splitting and completing the square to integrate.



3. Definite Integrals


Once you've mastered the technique of finding indefinite integrals using partial fractions, you will often apply it to evaluate definite integrals of rational functions. The process involves finding the antiderivative and then applying the limits of integration.

CBSE vs JEE: Both exams heavily test definite integrals, and partial fractions is a frequent technique used within them. JEE problems might involve more complex limits or properties of definite integrals alongside partial fractions.

4. Applications of Integrals (Area, Volume, Differential Equations)


The true utility of integration by partial fractions becomes evident in various application-based problems.


  • Area and Volume: Calculating areas under curves or volumes of solids of revolution where the integrand happens to be a rational function requiring partial fraction decomposition.


  • Differential Equations: Solving certain types of first-order differential equations, particularly those involving variable separation, often leads to integrals of rational functions that are solved using partial fractions. For example, in problems related to population growth, chemical reactions, or fluid dynamics, the rate equations might be rational functions.



Mastering these related topics will not only strengthen your understanding of integration by partial fractions but also enhance your overall problem-solving skills in Integral Calculus. Keep practicing!
⚠️ Common Exam Traps

Common Exam Traps in Integration by Partial Fractions


Integration by partial fractions is a powerful technique, but it's also prone to several common errors in exams. Being aware of these traps can save valuable marks. Let's look at the most frequent pitfalls:





  • Trap 1: Forgetting to Check for Improper Rational Functions

    Description: This is arguably the most common mistake. Partial fraction decomposition is strictly applicable only to proper rational functions (degree of numerator < degree of denominator). If the degree of the numerator is greater than or equal to the degree of the denominator, you must first perform long division to express it as a polynomial plus a proper rational function.


    Example: For an integral like $int frac{x^3 + x}{x^2 - 1} dx$, students often try to directly decompose $frac{x^3 + x}{x^2 - 1}$ into partial fractions, which is incorrect. First, perform long division: $frac{x^3 + x}{x^2 - 1} = x + frac{2x}{x^2 - 1}$. Then, apply partial fractions to $frac{2x}{x^2 - 1}$.


    JEE & CBSE Alert: This step is fundamental. Missing it guarantees an incorrect solution and is a primary differentiator in problem-solving ability.





  • Trap 2: Incorrect Partial Fraction Decomposition Forms

    Description: Using the wrong form for decomposition based on the denominator's factors is a frequent error. Each type of factor requires a specific partial fraction term:



    • Linear Non-repeated: For $(ax+b)$, use $frac{A}{ax+b}$.

    • Linear Repeated: For $(ax+b)^n$, use $frac{A_1}{ax+b} + frac{A_2}{(ax+b)^2} + dots + frac{A_n}{(ax+b)^n}$. Students often forget subsequent powers.

    • Irreducible Quadratic: For $(ax^2+bx+c)$ where $b^2-4ac < 0$, use $frac{Ax+B}{ax^2+bx+c}$. A common mistake is using just $frac{A}{ax^2+bx+c}$.

    • Repeated Irreducible Quadratic: For $(ax^2+bx+c)^n$, use $frac{A_1x+B_1}{ax^2+bx+c} + frac{A_2x+B_2}{(ax^2+bx+c)^2} + dots + frac{A_nx+B_n}{(ax^2+bx+c)^n}$. This is complex and often misapplied.


    JEE Alert: JEE often includes more complex denominators involving repeated linear or irreducible quadratic factors, requiring precise decomposition forms.





  • Trap 3: Algebraic Errors While Solving for Constants

    Description: After setting up the partial fraction decomposition, solving for the unknown constants (A, B, C, etc.) involves equating coefficients or substituting specific values of x. Algebraic mistakes in these steps are very common:



    • Incorrect expansion: Errors in multiplying out terms after finding a common denominator.

    • Equating coefficients: Mistakes in solving the system of linear equations obtained by comparing coefficients of powers of x.

    • Substitution method: While substituting roots of the denominator (e.g., $x=1$ if $(x-1)$ is a factor) can quickly find some constants, it's often not sufficient for all constants, especially with repeated or irreducible quadratic factors. Over-reliance or incorrect application of this method can lead to errors.





  • Trap 4: Errors in the Final Integration Step

    Description: Even if the decomposition and constants are correct, students sometimes make mistakes in integrating the resulting terms:



    • Logarithm terms: Forgetting the coefficient of x in the denominator, e.g., $int frac{1}{ax+b} dx = frac{1}{a} ln|ax+b| + C$, not just $ln|ax+b|$.

    • Inverse trigonometric terms: Incorrectly integrating terms like $frac{1}{x^2+a^2}$ or $frac{1}{(x-h)^2+a^2}$ (which integrate to $frac{1}{a} an^{-1}(frac{x}{a})$ or $frac{1}{a} an^{-1}(frac{x-h}{a})$ respectively). Completing the square might be necessary for quadratic denominators.

    • Power Rule: Incorrectly applying the power rule for terms like $int frac{1}{(ax+b)^n} dx$. Remember it's $int (ax+b)^{-n} dx = frac{(ax+b)^{-n+1}}{a(-n+1)} + C$.


    Forgetting the Constant of Integration (+C): A minor but crucial error that can cost marks, particularly in CBSE exams.






Key Takeaway: Always approach partial fraction problems systematically: Check for proper/improper, decompose carefully, solve for constants accurately, and finally, integrate each term meticulously. Practice with varied problems is key to avoiding these common traps.


⭐ Key Takeaways

Key Takeaways: Integration by Partial Fractions



Integration by Partial Fractions is a powerful technique used to integrate rational functions that cannot be easily integrated by substitution or other standard methods. It involves decomposing a complex rational function into simpler fractions, which are then integrated individually.



1. When to Use Partial Fractions



  • This method is applicable for integrating rational functions, i.e., functions of the form $frac{P(x)}{Q(x)}$, where $P(x)$ and $Q(x)$ are polynomials.

  • Crucial First Step: Proper Rational Function

    • If the degree of $P(x)$ is greater than or equal to the degree of $Q(x)$ (an improper rational function), always perform polynomial long division first. This yields $frac{P(x)}{Q(x)} = S(x) + frac{R(x)}{Q(x)}$, where $S(x)$ is the quotient and $R(x)$ is the remainder, and the degree of $R(x)$ is less than the degree of $Q(x)$. You then apply partial fractions to $frac{R(x)}{Q(x)}$.

    • If the degree of $P(x)$ is less than the degree of $Q(x)$ (a proper rational function), you can directly proceed with partial fraction decomposition.



  • The denominator $Q(x)$ must be factorizable into linear and/or irreducible quadratic factors.



2. Types of Partial Fraction Decomposition


The form of decomposition depends on the nature of the factors in the denominator $Q(x)$. Here's a summary of common cases:
































Factor Type in $Q(x)$ Corresponding Partial Fraction Form Example Decomposition
Linear Non-repeated: $(ax+b)$ $frac{A}{ax+b}$ $frac{P(x)}{(x-1)(x-2)} = frac{A}{x-1} + frac{B}{x-2}$
Linear Repeated: $(ax+b)^n$ $frac{A_1}{ax+b} + frac{A_2}{(ax+b)^2} + dots + frac{A_n}{(ax+b)^n}$ $frac{P(x)}{(x-1)^2(x-2)} = frac{A}{x-1} + frac{B}{(x-1)^2} + frac{C}{x-2}$
Irreducible Quadratic Non-repeated: $(ax^2+bx+c)$ (where $b^2-4ac < 0$) $frac{Ax+B}{ax^2+bx+c}$ $frac{P(x)}{(x-1)(x^2+4)} = frac{A}{x-1} + frac{Bx+C}{x^2+4}$
Irreducible Quadratic Repeated: $(ax^2+bx+c)^n$ (More common in JEE Advanced) $frac{A_1x+B_1}{ax^2+bx+c} + frac{A_2x+B_2}{(ax^2+bx+c)^2} + dots + frac{A_nx+B_n}{(ax^2+bx+c)^n}$ $frac{P(x)}{(x^2+1)^2} = frac{Ax+B}{x^2+1} + frac{Cx+D}{(x^2+1)^2}$


3. Determining the Constants



  • After setting up the decomposition, combine the partial fractions back into a single fraction with the original denominator.

  • Equate the numerator of this combined fraction with the original numerator $P(x)$.

  • Methods to find constants (A, B, C, etc.):

    • Substitution Method: For linear factors $(x-a)$, substitute $x=a$ into the equated numerator equation. This often directly yields one or more constants. This is usually the quickest way for distinct linear factors.

    • Equating Coefficients: Expand the right side and equate the coefficients of like powers of $x$ (e.g., $x^2, x^1, x^0$) on both sides of the numerator equation. This generates a system of linear equations, which can be solved for the constants. This method is essential for irreducible quadratic factors and often necessary for repeated factors or when direct substitution isn't sufficient.





4. Final Integration



  • Once the constants are found, each partial fraction is a standard integral.

    • $int frac{A}{ax+b} dx = frac{A}{a} ln|ax+b| + C$

    • $int frac{A}{(ax+b)^n} dx = frac{A}{a} frac{(ax+b)^{-n+1}}{-n+1} + C$ (for $n
      eq 1$      )

    • $int frac{Ax+B}{ax^2+bx+c} dx$ often involves splitting into two parts: one for a derivative of the denominator (leading to $ln$) and the other for $ an^{-1}$ form after completing the square in the denominator.





5. CBSE vs. JEE Main Perspective



  • CBSE: Focus primarily on linear (distinct and repeated) and simple irreducible quadratic factors. The problems are usually direct applications of the standard forms.

  • JEE Main: Expect more complex denominators, sometimes requiring clever factorization or a substitution before partial fractions. Irreducible quadratic factors (non-repeated) are common. Repeated irreducible quadratic factors are less common but can appear in challenging problems or advanced topics. Always look out for polynomial division first!



Mastering partial fractions requires keen observation of the denominator's factors and methodical algebraic manipulation. Practice is key!

🧩 Problem Solving Approach

Problem-Solving Approach: Integration by Partial Fractions



Integration by partial fractions is a powerful technique used to integrate rational functions, i.e., functions of the form P(x)/Q(x), where P(x) and Q(x) are polynomials. The core idea is to decompose the complex rational function into simpler rational functions that are easier to integrate.

When to Use Partial Fractions?


You should consider partial fraction decomposition when:

  • The integrand is a rational function of the form P(x)/Q(x).

  • The degree of the numerator P(x) is strictly less than the degree of the denominator Q(x). If not, perform polynomial long division first to convert it into a proper rational function plus a polynomial.

  • The denominator Q(x) can be factorized into linear and/or irreducible quadratic factors.



Systematic Steps for Integration by Partial Fractions


Follow these steps for a systematic approach to solving problems involving partial fractions:


  1. Step 1: Check Proper Fraction Condition

    • Compare the degree of P(x) with Q(x).

    • If deg(P(x)) $ge$ deg(Q(x)), perform polynomial long division:

      $frac{P(x)}{Q(x)} = S(x) + frac{R(x)}{Q(x)}$

      where S(x) is the quotient and R(x) is the remainder, with deg(R(x)) < deg(Q(x)). You will then integrate S(x) directly and apply partial fractions to $frac{R(x)}{Q(x)}$.

      JEE Specific: This step is crucial and more frequently tested in JEE.



  2. Step 2: Factorize the Denominator Q(x)

    • Factorize Q(x) completely into its constituent linear and/or irreducible quadratic factors.

    • Example: $x^3 - x = x(x^2-1) = x(x-1)(x+1)$



  3. Step 3: Decompose into Partial Fractions
    Based on the factors of Q(x), write down the appropriate partial fraction decomposition. This is the most critical step and depends on the nature of the factors:

    • Case 1: Non-repeated Linear Factors (e.g., $(x-a)(x-b)$)

      $frac{P(x)}{(x-a)(x-b)} = frac{A}{x-a} + frac{B}{x-b}$

    • Case 2: Repeated Linear Factors (e.g., $(x-a)^n$)

      $frac{P(x)}{(x-a)^n} = frac{A_1}{x-a} + frac{A_2}{(x-a)^2} + dots + frac{A_n}{(x-a)^n}$

    • Case 3: Non-repeated Irreducible Quadratic Factors (e.g., $(ax^2+bx+c)$ where $b^2-4ac < 0$)

      $frac{P(x)}{(ax^2+bx+c)(dx-e)} = frac{Ax+B}{ax^2+bx+c} + frac{C}{dx-e}$

      JEE Specific: This case is more common and elaborate in JEE problems compared to CBSE.

    • Case 4: Repeated Irreducible Quadratic Factors (e.g., $(ax^2+bx+c)^n$)

      $frac{P(x)}{(ax^2+bx+c)^n} = frac{A_1x+B_1}{ax^2+bx+c} + frac{A_2x+B_2}{(ax^2+bx+c)^2} + dots + frac{A_nx+B_n}{(ax^2+bx+c)^n}$

      JEE Specific: This is a more advanced case rarely seen in CBSE, but possible in JEE.



  4. Step 4: Determine the Constants

    • Multiply both sides of the partial fraction equation by the original denominator Q(x) to clear the denominators.

    • Equate the numerators.

    • Method 1: Substitution Method (Most efficient for linear factors): Substitute the roots of the linear factors (values of x that make a factor zero) into the equation to directly find some constants.

    • Method 2: Comparing Coefficients: Expand the right side and equate the coefficients of like powers of x (e.g., $x^2$, $x$, constant term) on both sides. Solve the resulting system of linear equations.

    • Often, a combination of both methods is the fastest.



  5. Step 5: Integrate Each Partial Fraction
    Replace the original integral with the sum of the integrals of the simpler partial fractions.

    • $int frac{A}{ax+b} dx = frac{A}{a} ln|ax+b| + C$

    • $int frac{A}{(ax+b)^n} dx = frac{A}{a(1-n)}(ax+b)^{1-n} + C$ (for $n
      e 1$)

    • Integrals involving irreducible quadratic factors often lead to $ln$ terms (by making the numerator derivative of denominator) and $arctan$ terms (by completing the square in the denominator).





Mastering these steps, especially the decomposition cases and the method for finding constants, is key to successfully applying this technique in exams.

📝 CBSE Focus Areas

CBSE Focus Areas: Integration by Partial Fractions


Integration by Partial Fractions is a crucial technique in Integral Calculus, frequently tested in CBSE Board Examinations. For board exams, the emphasis is on a systematic approach, clear presentation of steps, and mastery of standard decomposition forms.



Key Concepts for CBSE Boards:


The core idea is to decompose a rational function (a fraction where both numerator and denominator are polynomials) into simpler fractions that are easier to integrate. CBSE typically focuses on proper rational functions where the degree of the numerator is strictly less than the degree of the denominator. If it's an improper rational function (degree of numerator ≥ degree of denominator), polynomial long division must be performed first to convert it into a sum of a polynomial and a proper rational function.




  • Standard Forms of Decomposition: CBSE primarily focuses on three main types of denominators:

    1. Distinct Linear Factors: For a denominator like $(x-a)(x-b)$, the decomposition is $ frac{A}{x-a} + frac{B}{x-b} $.

    2. Repeated Linear Factors: For a denominator like $(x-a)^2(x-b)$, the decomposition is $ frac{A}{x-a} + frac{B}{(x-a)^2} + frac{C}{x-b} $.

    3. Irreducible Quadratic Factors: For a denominator like $(x-a)(x^2+bx+c)$ where $x^2+bx+c$ cannot be factored into real linear factors (i.e., its discriminant $b^2-4ac < 0$), the decomposition is $ frac{A}{x-a} + frac{Bx+C}{x^2+bx+c} $.



  • Methods to Find Constants (A, B, C...):

    • Substitution Method: Substitute roots of the linear factors into the original equation (after clearing denominators) to find corresponding constants quickly.

    • Comparing Coefficients: Expand the decomposed form, equate coefficients of like powers of 'x' on both sides, and solve the resulting system of linear equations. Often, a combination of both methods is most efficient.



  • Integration after Decomposition: After successfully decomposing the fraction, integration typically involves standard forms like $int frac{1}{x} dx = ln|x| + C$ or $int frac{1}{x^2+a^2} dx = frac{1}{a} an^{-1}left(frac{x}{a}
    ight) + C$. Sometimes, completing the square in the denominator of quadratic terms is required before integration.



Common Pitfalls & CBSE Exam Tips:



  • Polynomial Division: A frequent mistake is not performing polynomial division when the degree of the numerator is greater than or equal to the degree of the denominator. This is a mandatory first step for improper rational functions.

  • Identifying Correct Form: Carefully identify the nature of factors in the denominator (distinct linear, repeated linear, irreducible quadratic) to apply the correct partial fraction form.

  • Algebraic Errors: Be meticulous while solving for constants A, B, C. Simple algebraic mistakes can lead to incorrect final answers.

  • Constant of Integration (+C): Always remember to add the constant of integration '+C' for indefinite integrals. Missing this can lead to loss of marks in CBSE.

  • Show All Steps: For subjective CBSE exams, presenting each step of decomposition and integration clearly is crucial for full marks. Marks are often awarded for correct intermediate steps.



CBSE vs. JEE Main Perspective:


While JEE Main might involve more complex algebraic manipulations, higher-degree polynomials, or combinations with other advanced integration techniques, CBSE questions typically stick to straightforward applications of the standard partial fraction forms discussed above. The focus in CBSE is more on the procedural understanding and accurate execution of the method.




Mastering partial fractions for CBSE ensures you can confidently tackle these scoring questions with precision and clarity.


🎓 JEE Focus Areas

Integration by Partial Fractions: JEE Focus Areas



Integration by partial fractions is a fundamental technique in Integral Calculus, primarily used to integrate rational functions (a ratio of two polynomials). For JEE Main, a strong grasp of this method is crucial, as it often appears both directly and as a step in more complex problems involving definite integrals or area under curves.



The core idea is to decompose a complex rational function into simpler rational functions, which are then easier to integrate using standard formulas.



1. Proper vs. Improper Rational Functions - The First Critical Step




  • JEE Alert: Before attempting partial fractions, always check if the given rational function $frac{P(x)}{Q(x)}$ is proper or improper.


  • Proper Rational Function: Degree of numerator $P(x)$ is less than the degree of denominator $Q(x)$ (deg($P(x)$) < deg($Q(x)$)). This is the standard form for partial fraction decomposition.


  • Improper Rational Function: Degree of numerator $P(x)$ is greater than or equal to the degree of denominator $Q(x)$ (deg($P(x)$) $ge$ deg($Q(x)$)).

    • Action Required: If improper, perform polynomial long division first. This expresses $frac{P(x)}{Q(x)}$ as $S(x) + frac{R(x)}{Q(x)}$, where $S(x)$ is the quotient (a polynomial, easily integrable) and $frac{R(x)}{Q(x)}$ is a proper rational function (which then undergoes partial fraction decomposition).

    • Common JEE Mistake: Forgetting long division for improper fractions.





2. Mastering the Standard Decomposition Forms


The form of decomposition depends entirely on the factors of the denominator $Q(x)$. JEE often tests combinations of these cases.

































Type of Factors in $Q(x)$ Form of Partial Fraction Example
1. Non-repeated linear factors:
$(ax+b)(cx+d)$		 $frac{A}{ax+b} + frac{B}{cx+d}$ $frac{x+1}{(x-2)(x+3)} = frac{A}{x-2} + frac{B}{x+3}$
2. Repeated linear factors:
$(ax+b)^n$		 $frac{A_1}{ax+b} + frac{A_2}{(ax+b)^2} + dots + frac{A_n}{(ax+b)^n}$ $frac{1}{(x-1)^2(x+2)} = frac{A}{x-1} + frac{B}{(x-1)^2} + frac{C}{x+2}$
3. Non-repeated irreducible quadratic factors:
$(ax^2+bx+c)$ (where $b^2-4ac < 0$)		 $frac{Ax+B}{ax^2+bx+c}$ $frac{2x+1}{(x^2+1)(x-1)} = frac{Ax+B}{x^2+1} + frac{C}{x-1}$
4. Repeated irreducible quadratic factors:
$(ax^2+bx+c)^n$		 $frac{A_1x+B_1}{ax^2+bx+c} + dots + frac{A_nx+B_n}{(ax^2+bx+c)^n}$ $frac{x^2+1}{(x^2+x+1)^2} = frac{Ax+B}{x^2+x+1} + frac{Cx+D}{(x^2+x+1)^2}$


3. Efficient Determination of Coefficients (JEE Speed Techniques)




  • Heaviside's Cover-up Method: Highly recommended for JEE when dealing with non-repeated linear factors. To find a coefficient for a factor $(x-a)$, cover up $(x-a)$ in the original fraction and substitute $x=a$ into the remaining expression. This is exceptionally fast.


  • Substitution Method: Substitute convenient values of $x$ (roots of the factors, or $x=0, 1, -1$) into the partially decomposed equation to quickly find coefficients.


  • Comparing Coefficients: Equating coefficients of like powers of $x$ on both sides of the equation. This is a general method, but can be time-consuming for higher degree polynomials. Use it when other methods are not applicable or insufficient (e.g., for quadratic factors or finding all coefficients in repeated linear factors after initial ones).



4. Post-Decomposition Integration Forms


After partial fraction decomposition, you will typically encounter integrals of the form:



  • $int frac{A}{ax+b} dx = frac{A}{a} ln|ax+b| + C$

  • $int frac{A}{(ax+b)^n} dx = frac{A}{a} frac{(ax+b)^{-n+1}}{-n+1} + C$ (for $n
    e 1$)

  • $int frac{Ax+B}{ax^2+bx+c} dx$: Requires splitting into $int frac{A' (2ax+b)}{ax^2+bx+c} dx$ (for $ln$) and $int frac{B'}{ax^2+bx+c} dx$ (for $ an^{-1}$, completing the square in the denominator). This is a common pattern in JEE.



JEE vs. CBSE Perspective




  • CBSE: Typically focuses on simpler cases (non-repeated and repeated linear factors), with straightforward coefficient determination. The integration steps are often basic.


  • JEE: Expect more complex denominators, combinations of all factor types, and sometimes implicit applications within definite integrals or problems involving areas. Speed in decomposition and accurate integration of quadratic terms are vital. Questions might involve finding the values of A, B, C directly without requiring full integration.



Key Takeaway for JEE: Master the decomposition forms and develop efficiency in finding coefficients using smart techniques like Heaviside's method. Always remember the proper/improper fraction check.

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Status: AI Generated
Version: 1
Generated: Oct 22, 2025
🌐 Overview
Integration by partial fractions decomposes a rational function P(x)/Q(x) (deg P < deg Q) into a sum of simpler fractions with linear or irreducible quadratic denominators. Standard cases: distinct linear factors, repeated linear factors, and irreducible quadratic factors (with linear numerators). After decomposition, integrate term-wise using basic formulas.
📚 Fundamentals
• Distinct linear: A1/(x−a1)+…+An/(x−an).
• Repeated linear: A1/(x−a)+A2/(x−a)^2+…
• Irreducible quadratic: (Ax+B)/(x^2+px+q).
• Integrals: logs for linear terms; arctan/log mix for quadratic terms after completing square.
🔬 Deep Dive
Heaviside’s method justification (distinct roots), relation to Laplace transform inversions, and partial fractions in complex analysis (overview).
🎯 Shortcuts
“Divide–Factor–Decompose–Integrate” (DFDI).
💡 Quick Tips
• For distinct linear factors, the cover-up method speeds up A_i.
• For quadratics, complete the square early to see arctan structure.
• Keep constants outside and simplify before integrating.
🧠 Intuitive Understanding
Break a complex rational function into “bite-sized” pieces whose integrals are well-known (logs and arctans).
🌍 Real World Applications
• Control systems and Laplace transforms.
• Probability generating functions and partial fraction expansion.
• Circuit analysis and differential equations solutions.
🔄 Common Analogies
• Factoring a number into primes—factor the denominator and match numerators accordingly.
📋 Prerequisites
Polynomial division (if deg P ≥ deg Q), factoring polynomials, solving linear systems for coefficients, standard integrals for logs/arctans.
🔗 Related Topics
Integration by substitution, completing the square for quadratics, Heaviside cover-up method (for simple linear distinct factors).
⚠️ Common Exam Traps
• Forgetting to divide when deg P ≥ deg Q.
• Wrong template (e.g., treating quadratic as linear).
• Arithmetic slips in coefficients leading to bad integrals.
⭐ Key Takeaways
• Always reduce degree first by long division.
• Choose the decomposition type matching factorization.
• Solve coefficients carefully; check by recombining.
🧩 Problem Solving Approach
1) Divide if needed.
2) Factor Q; write decomposition.
3) Determine coefficients; integrate term-wise.
4) Differentiate final answer to verify if needed.
📝 CBSE Focus Areas
Standard templates, quick coefficient solving, and clean term-by-term integration.
🎓 JEE Focus Areas
Speed in factoring and coefficient determination; mixing substitution with partial fractions; handling repeated/irreducible cases under time pressure.

📊 AI Generation Metadata

Estimated Time: 90 mins
AI Model: ChatGPT 5 (Copilot Agent)
Status: AI Generated
Version: 1
Generated: Oct 23, 2025

📝CBSE 12th Board Problems (18)

Problem 255
Medium 5 Marks
Evaluate the integral (int frac{x^2+x+1}{x^2-x} dx).
Show Solution
1. Perform long division because the degree of the numerator is equal to the degree of the denominator (improper rational function). (frac{x^2+x+1}{x^2-x} = frac{(x^2-x) + 2x+1}{x^2-x} = 1 + frac{2x+1}{x^2-x}). 2. Focus on integrating (frac{2x+1}{x^2-x}). Factor the denominator: (x^2-x = x(x-1)). 3. Decompose into partial fractions: (frac{2x+1}{x(x-1)} = frac{A}{x} + frac{B}{x-1}). 4. Solve for A and B: (2x+1 = A(x-1) + Bx). - Set x = 0: (1 = A(-1) implies A = -1). - Set x = 1: (3 = B(1) implies B = 3). 5. Substitute A and B: (frac{-1}{x} + frac{3}{x-1}). 6. Now integrate the original expression: (int left( 1 + frac{-1}{x} + frac{3}{x-1} ight) dx). 7. Evaluate the integral: (x - log|x| + 3log|x-1| + C). 8. Simplify: (x + logleft|frac{(x-1)^3}{x} ight| + C).
Final Answer: (x + logleft|frac{(x-1)^3}{x} ight| + C)
Problem 255
Hard 6 Marks
Integrate the function $frac{1}{x(x^n+1)}$ with respect to $x$.
Show Solution
1. **Modify the integrand for substitution:** To make a substitution easier, multiply the numerator and denominator by $x^{n-1}$: $$int frac{x^{n-1}}{x^n(x^n+1)} dx$$ 2. **Perform substitution:** Let $t = x^n$. Then $dt = nx^{n-1} dx implies x^{n-1} dx = frac{1}{n} dt$. The integral becomes: $$int frac{1}{n} frac{dt}{t(t+1)}$$ 3. **Set up partial fractions:** The denominator has two distinct linear factors. $$frac{1}{t(t+1)} = frac{A}{t} + frac{B}{t+1}$$ 4. **Solve for constants:** Multiply by $t(t+1)$: $1 = A(t+1) + Bt$ * Set $t=0$: $1 = A(1) implies A=1$. * Set $t=-1$: $1 = B(-1) implies B=-1$. 5. **Substitute constants and integrate (in terms of t):** $$frac{1}{n} int left(frac{1}{t} - frac{1}{t+1} ight) dt$$ $$= frac{1}{n} (log|t| - log|t+1|) + C$$ $$= frac{1}{n} logleft|frac{t}{t+1} ight| + C$$ 6. **Substitute back $t=x^n$:** $$= frac{1}{n} logleft|frac{x^n}{x^n+1} ight| + C$$
Final Answer: $frac{1}{n} logleft|frac{x^n}{x^n+1} ight| + C$
Problem 255
Hard 6 Marks
Evaluate $int frac{x^2+1}{(x-1)(x-2)(x-3)} dx$
Show Solution
1. **Set up partial fractions:** The denominator has three distinct linear factors. $$frac{x^2+1}{(x-1)(x-2)(x-3)} = frac{A}{x-1} + frac{B}{x-2} + frac{C}{x-3}$$ 2. **Solve for constants:** Multiply by $(x-1)(x-2)(x-3)$: $x^2+1 = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)$ * Set $x=1$: $1^2+1 = A(1-2)(1-3) + B(0) + C(0) implies 2 = A(-1)(-2) implies 2 = 2A implies A = 1$. * Set $x=2$: $2^2+1 = A(0) + B(2-1)(2-3) + C(0) implies 5 = B(1)(-1) implies 5 = -B implies B = -5$. * Set $x=3$: $3^2+1 = A(0) + B(0) + C(3-1)(3-2) implies 10 = C(2)(1) implies 10 = 2C implies C = 5$. 3. **Substitute constants and integrate:** $$int left(frac{1}{x-1} + frac{-5}{x-2} + frac{5}{x-3} ight) dx$$ $$= int frac{1}{x-1} dx - 5int frac{1}{x-2} dx + 5int frac{1}{x-3} dx$$ $$= log|x-1| - 5log|x-2| + 5log|x-3| + C$$ 4. **Combine logarithmic terms:** $$= log|x-1| + 5(log|x-3| - log|x-2|) + C$$ $$= log|x-1| + 5logleft|frac{x-3}{x-2} ight| + C$$
Final Answer: $log|x-1| - 5log|x-2| + 5log|x-3| + C$ or $logleft|x-1left(frac{x-3}{x-2} ight)^5 ight| + C$
Problem 255
Hard 4 Marks
Evaluate $int frac{e^x}{(e^x-1)(e^x+2)} dx$
Show Solution
1. **Perform substitution:** Let $t = e^x$. Then $dt = e^x dx$. The integral becomes: $$int frac{dt}{(t-1)(t+2)}$$ 2. **Set up partial fractions:** The denominator has two distinct linear factors. $$frac{1}{(t-1)(t+2)} = frac{A}{t-1} + frac{B}{t+2}$$ 3. **Solve for constants:** Multiply by $(t-1)(t+2)$: $1 = A(t+2) + B(t-1)$ * Set $t=1$: $1 = A(1+2) + B(0) implies 1 = 3A implies A = 1/3$. * Set $t=-2$: $1 = A(0) + B(-2-1) implies 1 = -3B implies B = -1/3$. 4. **Substitute constants and integrate (in terms of t):** $$int left(frac{1/3}{t-1} + frac{-1/3}{t+2} ight) dt$$ $$= frac{1}{3}int frac{1}{t-1} dt - frac{1}{3}int frac{1}{t+2} dt$$ $$= frac{1}{3}log|t-1| - frac{1}{3}log|t+2| + C$$ 5. **Substitute back $t=e^x$ and combine terms:** $$= frac{1}{3}(log|e^x-1| - log|e^x+2|) + C$$ $$= frac{1}{3}logleft|frac{e^x-1}{e^x+2} ight| + C$$
Final Answer: $frac{1}{3}logleft|frac{e^x-1}{e^x+2} ight| + C$
Problem 255
Hard 6 Marks
Find the integral of $frac{x^2}{(x^2+1)(x-1)}$ with respect to $x$.
Show Solution
1. **Check for proper fraction:** The degree of the numerator (2) is less than the degree of the denominator (3), so it's a proper fraction. 2. **Set up partial fractions:** The denominator has an irreducible quadratic factor $(x^2+1)$ and a distinct linear factor $(x-1)$. The decomposition is: $$frac{x^2}{(x^2+1)(x-1)} = frac{Ax+B}{x^2+1} + frac{C}{x-1}$$ 3. **Solve for constants:** Multiply by $(x^2+1)(x-1)$: $x^2 = (Ax+B)(x-1) + C(x^2+1)$ * Set $x=1$: $1^2 = (A(1)+B)(0) + C(1^2+1) implies 1 = 2C implies C = 1/2$. * Set $x=0$: $0^2 = (A(0)+B)(0-1) + C(0^2+1) implies 0 = B(-1) + C(1) implies 0 = -B + C implies B = C = 1/2$. * Compare coefficient of $x^2$: $x^2 = Ax^2 - Ax + Bx - B + Cx^2 + C$ $1 = A + C$. Since $C=1/2$, $1 = A + 1/2 implies A = 1/2$. 4. **Substitute constants and integrate:** $$int left(frac{frac{1}{2}x + frac{1}{2}}{x^2+1} + frac{frac{1}{2}}{x-1} ight) dx$$ $$= frac{1}{2}int frac{x+1}{x^2+1} dx + frac{1}{2}int frac{1}{x-1} dx$$ $$= frac{1}{2}int frac{x}{x^2+1} dx + frac{1}{2}int frac{1}{x^2+1} dx + frac{1}{2}int frac{1}{x-1} dx$$ * For $int frac{x}{x^2+1} dx$, let $u = x^2+1, du = 2x dx implies x dx = frac{1}{2} du$. Integral becomes $frac{1}{2}int frac{1}{u} du = frac{1}{2}log|u| = frac{1}{2}log(x^2+1)$. * For $int frac{1}{x^2+1} dx$, it's $ an^{-1}x$. * For $int frac{1}{x-1} dx$, it's $log|x-1|$. 5. **Combine results:** $$= frac{1}{2}left(frac{1}{2}log(x^2+1) ight) + frac{1}{2} an^{-1}x + frac{1}{2}log|x-1| + C$$ $$= frac{1}{4}log(x^2+1) + frac{1}{2} an^{-1}x + frac{1}{2}log|x-1| + C$$
Final Answer: $frac{1}{4}log(x^2+1) + frac{1}{2} an^{-1}x + frac{1}{2}log|x-1| + C$
Problem 255
Hard 6 Marks
Integrate: $int frac{dx}{(x-1)^2(x+2)}$
Show Solution
1. **Set up partial fractions:** The denominator has a repeated linear factor $(x-1)^2$ and a distinct linear factor $(x+2)$. The partial fraction decomposition is: $$frac{1}{(x-1)^2(x+2)} = frac{A}{x-1} + frac{B}{(x-1)^2} + frac{C}{x+2}$$ 2. **Solve for constants:** Multiply by $(x-1)^2(x+2)$: $1 = A(x-1)(x+2) + B(x+2) + C(x-1)^2$ * Set $x=1$: $1 = A(0) + B(1+2) + C(0) implies 1 = 3B implies B = 1/3$. * Set $x=-2$: $1 = A(0) + B(0) + C(-2-1)^2 implies 1 = C(-3)^2 implies 1 = 9C implies C = 1/9$. * To find A, compare coefficients or substitute another value, e.g., $x=0$: $1 = A(-1)(2) + B(2) + C(-1)^2$ $1 = -2A + 2B + C$ $1 = -2A + 2(1/3) + 1/9$ $1 = -2A + 2/3 + 1/9$ $1 = -2A + 6/9 + 1/9 = -2A + 7/9$ $1 - 7/9 = -2A implies 2/9 = -2A implies A = -1/9$. 3. **Substitute constants and integrate:** $$int left(frac{-1/9}{x-1} + frac{1/3}{(x-1)^2} + frac{1/9}{x+2} ight) dx$$ $$= -frac{1}{9}int frac{1}{x-1} dx + frac{1}{3}int (x-1)^{-2} dx + frac{1}{9}int frac{1}{x+2} dx$$ $$= -frac{1}{9}log|x-1| + frac{1}{3} frac{(x-1)^{-1}}{-1} + frac{1}{9}log|x+2| + C$$ $$= -frac{1}{9}log|x-1| - frac{1}{3(x-1)} + frac{1}{9}log|x+2| + C$$ 4. **Combine logarithmic terms:** $$= frac{1}{9}logleft|frac{x+2}{x-1} ight| - frac{1}{3(x-1)} + C$$
Final Answer: $frac{1}{9}logleft|frac{x+2}{x-1} ight| - frac{1}{3(x-1)} + C$
Problem 255
Hard 6 Marks
Evaluate the integral: $int frac{x^3 - x + 1}{x^2 + x} dx$
Show Solution
1. **Check for improper fraction:** The degree of the numerator (3) is greater than the degree of the denominator (2), so it's an improper fraction. Perform polynomial long division: $(x^3 - x + 1) = x(x^2 + x) - x^2 - x + 1$ $= x(x^2 + x) - (x^2 + x) + 1$ So, $frac{x^3 - x + 1}{x^2 + x} = x - 1 + frac{1}{x^2 + x}$. 2. **Factor the denominator:** $x^2 + x = x(x+1)$. 3. **Apply partial fractions to the remainder:** Let $frac{1}{x(x+1)} = frac{A}{x} + frac{B}{x+1}$. $1 = A(x+1) + Bx$. Setting $x=0 implies 1 = A(1) implies A=1$. Setting $x=-1 implies 1 = B(-1) implies B=-1$. So, $frac{1}{x(x+1)} = frac{1}{x} - frac{1}{x+1}$. 4. **Integrate each term:** $int (x - 1 + frac{1}{x} - frac{1}{x+1}) dx$ $= int x dx - int 1 dx + int frac{1}{x} dx - int frac{1}{x+1} dx$ $= frac{x^2}{2} - x + log|x| - log|x+1| + C$. 5. **Combine logarithmic terms:** $= frac{x^2}{2} - x + log|frac{x}{x+1}| + C$.
Final Answer: $frac{x^2}{2} - x + log|frac{x}{x+1}| + C$
Problem 255
Medium 6 Marks
Evaluate the integral (int frac{1}{(x+1)(x-2)(x+3)} dx).
Show Solution
1. Decompose the rational function into partial fractions: (frac{1}{(x+1)(x-2)(x+3)} = frac{A}{x+1} + frac{B}{x-2} + frac{C}{x+3}). 2. Multiply by ((x+1)(x-2)(x+3)): (1 = A(x-2)(x+3) + B(x+1)(x+3) + C(x+1)(x-2)). 3. Solve for A, B, C: - Set x = -1: (1 = A(-1-2)(-1+3) + B(0) + C(0) implies 1 = A(-3)(2) implies 1 = -6A implies A = -1/6). - Set x = 2: (1 = A(0) + B(2+1)(2+3) + C(0) implies 1 = B(3)(5) implies 1 = 15B implies B = 1/15). - Set x = -3: (1 = A(0) + B(0) + C(-3+1)(-3-2) implies 1 = C(-2)(-5) implies 1 = 10C implies C = 1/10). 4. Substitute A, B, C: (frac{-1/6}{x+1} + frac{1/15}{x-2} + frac{1/10}{x+3}). 5. Integrate each term: (int left( frac{-1/6}{x+1} + frac{1/15}{x-2} + frac{1/10}{x+3} ight) dx). 6. Evaluate: (-frac{1}{6}log|x+1| + frac{1}{15}log|x-2| + frac{1}{10}log|x+3| + C).
Final Answer: (-frac{1}{6}log|x+1| + frac{1}{15}log|x-2| + frac{1}{10}log|x+3| + C)
Problem 255
Medium 4 Marks
Find the integral of (frac{cos x}{(1+sin x)(2+sin x)}) with respect to x.
Show Solution
1. Use substitution: Let (t = sin x). Then (dt = cos x dx). 2. The integral transforms to: (int frac{dt}{(1+t)(2+t)}). 3. Decompose the rational function into partial fractions: (frac{1}{(1+t)(2+t)} = frac{A}{1+t} + frac{B}{2+t}). 4. Solve for A and B: (1 = A(2+t) + B(1+t)). - Set t = -1: (1 = A(2-1) + B(0) implies 1 = A implies A = 1). - Set t = -2: (1 = A(0) + B(1-2) implies 1 = -B implies B = -1). 5. Substitute A and B: (frac{1}{1+t} + frac{-1}{2+t}). 6. Integrate with respect to t: (int left( frac{1}{1+t} - frac{1}{2+t} ight) dt = log|1+t| - log|2+t| + C). 7. Simplify using logarithm properties: (logleft|frac{1+t}{2+t} ight| + C). 8. Substitute back (t = sin x): (logleft|frac{1+sin x}{2+sin x} ight| + C).
Final Answer: (logleft|frac{1+sin x}{2+sin x} ight| + C)
Problem 255
Easy 3 Marks
Evaluate the integral: $int frac{1}{x^2 - 9} dx$
Show Solution
Let the integral be $I = int frac{1}{x^2 - 9} dx$. We can write the integrand as $frac{1}{x^2 - 3^2} = frac{1}{(x-3)(x+3)}$. Using partial fractions, let $frac{1}{(x-3)(x+3)} = frac{A}{x-3} + frac{B}{x+3}$. Multiplying by $(x-3)(x+3)$, we get $1 = A(x+3) + B(x-3)$. Set $x=3 Rightarrow 1 = A(6) Rightarrow A = frac{1}{6}$. Set $x=-3 Rightarrow 1 = B(-6) Rightarrow B = -frac{1}{6}$. So, $I = int left( frac{1/6}{x-3} - frac{1/6}{x+3} ight) dx$. $I = frac{1}{6} int frac{1}{x-3} dx - frac{1}{6} int frac{1}{x+3} dx$. $I = frac{1}{6} log|x-3| - frac{1}{6} log|x+3| + C$. $I = frac{1}{6} logleft|frac{x-3}{x+3} ight| + C$.
Final Answer: $frac{1}{6} logleft|frac{x-3}{x+3} ight| + C$
Problem 255
Medium 5 Marks
Integrate (frac{x}{(x-1)(x^2+1)}) with respect to x.
Show Solution
1. Decompose into partial fractions: (frac{x}{(x-1)(x^2+1)} = frac{A}{x-1} + frac{Bx+C}{x^2+1}). 2. Multiply by ((x-1)(x^2+1)): (x = A(x^2+1) + (Bx+C)(x-1)). 3. Solve for A, B, C: - Set x = 1: (1 = A(1^2+1) + (B(1)+C)(1-1) implies 1 = 2A implies A = 1/2). - Expand the equation: (x = Ax^2+A + Bx^2-Bx+Cx-C = (A+B)x^2 + (C-B)x + (A-C)). - Compare coefficients: - (x^2) coefficient: (0 = A+B implies B = -A = -1/2). - Constant term: (0 = A-C implies C = A = 1/2). 4. Substitute A, B, C: (frac{1/2}{x-1} + frac{(-1/2)x+1/2}{x^2+1} = frac{1}{2(x-1)} + frac{1}{2}frac{-x+1}{x^2+1}). 5. Integrate each term: (int left( frac{1}{2(x-1)} - frac{x}{2(x^2+1)} + frac{1}{2(x^2+1)} ight) dx). 6. Evaluate: - (int frac{1}{2(x-1)} dx = frac{1}{2}log|x-1|). - For (int frac{-x}{2(x^2+1)} dx), let (u = x^2+1 implies du = 2x dx). So (int frac{-1}{4u} du = -frac{1}{4}log|u| = -frac{1}{4}log(x^2+1)). - (int frac{1}{2(x^2+1)} dx = frac{1}{2} an^{-1}x). 7. Combine: (frac{1}{2}log|x-1| - frac{1}{4}log(x^2+1) + frac{1}{2} an^{-1}x + C).
Final Answer: (frac{1}{2}log|x-1| - frac{1}{4}log(x^2+1) + frac{1}{2} an^{-1}x + C)
Problem 255
Medium 4 Marks
Find the integral of (frac{3x-2}{(x+1)^2(x+3)}).
Show Solution
1. Decompose into partial fractions: (frac{3x-2}{(x+1)^2(x+3)} = frac{A}{x+1} + frac{B}{(x+1)^2} + frac{C}{x+3}). 2. Multiply by ((x+1)^2(x+3)): (3x-2 = A(x+1)(x+3) + B(x+3) + C(x+1)^2). 3. Solve for A, B, C: - Set x = -1: (3(-1)-2 = A(0) + B(-1+3) + C(0) implies -5 = 2B implies B = -5/2). - Set x = -3: (3(-3)-2 = A(0) + B(0) + C(-3+1)^2 implies -11 = C(-2)^2 implies -11 = 4C implies C = -11/4). - Compare coefficients of (x^2) (or set x=0): (0 = A + C implies A = -C implies A = 11/4). 4. Substitute A, B, C: (frac{11/4}{x+1} + frac{-5/2}{(x+1)^2} + frac{-11/4}{x+3}). 5. Integrate each term: (int left( frac{11/4}{x+1} - frac{5/2}{(x+1)^2} - frac{11/4}{x+3} ight) dx). 6. Evaluate: (frac{11}{4}log|x+1| - frac{5}{2} left( -frac{1}{x+1} ight) - frac{11}{4}log|x+3| + C). 7. Simplify: (frac{11}{4}logleft|frac{x+1}{x+3} ight| + frac{5}{2(x+1)} + C).
Final Answer: (frac{11}{4}logleft|frac{x+1}{x+3} ight| + frac{5}{2(x+1)} + C)
Problem 255
Medium 4 Marks
Integrate the function (frac{x}{(x+1)(x+2)}) with respect to x.
Show Solution
1. Decompose the rational function into partial fractions: (frac{x}{(x+1)(x+2)} = frac{A}{x+1} + frac{B}{x+2}). 2. Solve for A and B. Multiply by (x+1)(x+2) to get: (x = A(x+2) + B(x+1)). - Set x = -1: (-1 = A(-1+2) + B(-1+1) implies -1 = A implies A = -1). - Set x = -2: (-2 = A(-2+2) + B(-2+1) implies -2 = -B implies B = 2). 3. Substitute A and B back into the partial fraction form: (frac{-1}{x+1} + frac{2}{x+2}). 4. Integrate each term separately: (int left( frac{-1}{x+1} + frac{2}{x+2} ight) dx = -int frac{1}{x+1} dx + 2int frac{1}{x+2} dx). 5. Evaluate the integrals: (- log|x+1| + 2 log|x+2| + C). 6. Simplify using logarithm properties: (log left|frac{(x+2)^2}{x+1} ight| + C).
Final Answer: (log left|frac{(x+2)^2}{x+1} ight| + C)
Problem 255
Easy 4 Marks
Evaluate $int frac{e^x}{e^{2x}-4} dx$.
Show Solution
Let the integral be $I = int frac{e^x}{e^{2x}-4} dx$. Let $t = e^x$. Then $dt = e^x dx$. The integral transforms to $I = int frac{dt}{t^2-4}$. We can write the integrand as $frac{1}{t^2 - 2^2} = frac{1}{(t-2)(t+2)}$. Using partial fractions, let $frac{1}{(t-2)(t+2)} = frac{A}{t-2} + frac{B}{t+2}$. Multiplying by $(t-2)(t+2)$, we get $1 = A(t+2) + B(t-2)$. Set $t=2 Rightarrow 1 = A(2+2) Rightarrow 1 = 4A Rightarrow A = frac{1}{4}$. Set $t=-2 Rightarrow 1 = B(-2-2) Rightarrow 1 = -4B Rightarrow B = -frac{1}{4}$. So, $I = int left( frac{1/4}{t-2} - frac{1/4}{t+2} ight) dt$. $I = frac{1}{4} int frac{1}{t-2} dt - frac{1}{4} int frac{1}{t+2} dt$. $I = frac{1}{4} log|t-2| - frac{1}{4} log|t+2| + C$. Substitute back $t = e^x$: $I = frac{1}{4} log|e^x-2| - frac{1}{4} log|e^x+2| + C$. $I = frac{1}{4} logleft|frac{e^x-2}{e^x+2} ight| + C$.
Final Answer: $frac{1}{4} logleft|frac{e^x-2}{e^x+2} ight| + C$
Problem 255
Easy 4 Marks
Find $int frac{x^2+1}{x^2-1} dx$.
Show Solution
Let the integral be $I = int frac{x^2+1}{x^2-1} dx$. The degree of the numerator is equal to the degree of the denominator, so we perform long division first: $frac{x^2+1}{x^2-1} = frac{(x^2-1)+2}{x^2-1} = 1 + frac{2}{x^2-1}$. So, $I = int left( 1 + frac{2}{x^2-1} ight) dx = int 1 dx + 2 int frac{1}{x^2-1} dx$. We know that $int frac{1}{x^2-a^2} dx = frac{1}{2a} logleft|frac{x-a}{x+a} ight| + C$. Here $a=1$. So, $I = x + 2 left( frac{1}{2(1)} logleft|frac{x-1}{x+1} ight| ight) + C$. $I = x + logleft|frac{x-1}{x+1} ight| + C$.
Final Answer: $x + logleft|frac{x-1}{x+1} ight| + C$
Problem 255
Easy 4 Marks
Integrate $frac{1}{(x+1)^2(x+2)}$ with respect to $x$.
Show Solution
Let the integral be $I = int frac{1}{(x+1)^2(x+2)} dx$. Using partial fractions, let $frac{1}{(x+1)^2(x+2)} = frac{A}{x+1} + frac{B}{(x+1)^2} + frac{C}{x+2}$. Multiplying by $(x+1)^2(x+2)$, we get $1 = A(x+1)(x+2) + B(x+2) + C(x+1)^2$. Set $x=-1 Rightarrow 1 = A(0) + B(-1+2) + C(0) Rightarrow 1 = B(1) Rightarrow B = 1$. Set $x=-2 Rightarrow 1 = A(0) + B(0) + C(-2+1)^2 Rightarrow 1 = C(-1)^2 Rightarrow C = 1$. To find A, compare coefficients or substitute another value for x. Let $x=0$: $1 = A(1)(2) + B(2) + C(1)^2$. $1 = 2A + 2B + C$. Substitute $B=1$ and $C=1$: $1 = 2A + 2(1) + 1 Rightarrow 1 = 2A + 3 Rightarrow 2A = -2 Rightarrow A = -1$. So, $I = int left( frac{-1}{x+1} + frac{1}{(x+1)^2} + frac{1}{x+2} ight) dx$. $I = -int frac{1}{x+1} dx + int (x+1)^{-2} dx + int frac{1}{x+2} dx$. $I = -log|x+1| + frac{(x+1)^{-1}}{-1} + log|x+2| + C$. $I = -log|x+1| - frac{1}{x+1} + log|x+2| + C$. $I = logleft|frac{x+2}{x+1} ight| - frac{1}{x+1} + C$.
Final Answer: $logleft|frac{x+2}{x+1} ight| - frac{1}{x+1} + C$
Problem 255
Easy 3 Marks
Evaluate $int frac{3x-2}{(x+1)(x-2)} dx$.
Show Solution
Let the integral be $I = int frac{3x-2}{(x+1)(x-2)} dx$. Using partial fractions, let $frac{3x-2}{(x+1)(x-2)} = frac{A}{x+1} + frac{B}{x-2}$. Multiplying by $(x+1)(x-2)$, we get $3x-2 = A(x-2) + B(x+1)$. Set $x=-1 Rightarrow 3(-1)-2 = A(-1-2) Rightarrow -5 = A(-3) Rightarrow A = frac{5}{3}$. Set $x=2 Rightarrow 3(2)-2 = B(2+1) Rightarrow 4 = B(3) Rightarrow B = frac{4}{3}$. So, $I = int left( frac{5/3}{x+1} + frac{4/3}{x-2} ight) dx$. $I = frac{5}{3} int frac{1}{x+1} dx + frac{4}{3} int frac{1}{x-2} dx$. $I = frac{5}{3} log|x+1| + frac{4}{3} log|x-2| + C$.
Final Answer: $frac{5}{3} log|x+1| + frac{4}{3} log|x-2| + C$
Problem 255
Easy 3 Marks
Find the integral of $frac{x}{(x+1)(x+2)}$ with respect to $x$.
Show Solution
Let the integral be $I = int frac{x}{(x+1)(x+2)} dx$. Using partial fractions, let $frac{x}{(x+1)(x+2)} = frac{A}{x+1} + frac{B}{x+2}$. Multiplying by $(x+1)(x+2)$, we get $x = A(x+2) + B(x+1)$. Set $x=-1 Rightarrow -1 = A(-1+2) Rightarrow -1 = A(1) Rightarrow A = -1$. Set $x=-2 Rightarrow -2 = B(-2+1) Rightarrow -2 = B(-1) Rightarrow B = 2$. So, $I = int left( frac{-1}{x+1} + frac{2}{x+2} ight) dx$. $I = -int frac{1}{x+1} dx + 2int frac{1}{x+2} dx$. $I = -log|x+1| + 2log|x+2| + C$. $I = logleft|frac{(x+2)^2}{x+1} ight| + C$.
Final Answer: $2log|x+2| - log|x+1| + C$ or $logleft|frac{(x+2)^2}{x+1} ight| + C$

🎯IIT-JEE Main Problems (19)

Problem 255
Medium 4 Marks
Calculate the integral: (int frac{x^3+x^2+1}{x^2-1} dx)
Show Solution
1. Perform polynomial long division since the degree of the numerator is greater than or equal to the degree of the denominator (improper fraction).<br>&nbsp;&nbsp;&nbsp;&nbsp;Divide (x^3+x^2+1) by (x^2-1):<br>&nbsp;&nbsp;&nbsp;&nbsp;(x^3+x^2+1 = x(x^2-1) + x^2+x+1).<br>&nbsp;&nbsp;&nbsp;&nbsp;(x^2+x+1 = 1(x^2-1) + x+2).<br>&nbsp;&nbsp;&nbsp;&nbsp;So, (frac{x^3+x^2+1}{x^2-1} = x+1 + frac{x+2}{x^2-1}).<br>2. Factorize the remaining denominator: (x^2-1 = (x-1)(x+1)).<br>3. Decompose the rational part using partial fractions: Let (frac{x+2}{(x-1)(x+1)} = frac{A}{x-1} + frac{B}{x+1}).<br>4. Equate numerators: (x+2 = A(x+1) + B(x-1)).<br>5. Solve for A and B:<br>&nbsp;&nbsp;&nbsp;&nbsp;Set (x=1): (1+2 = A(1+1) + B(1-1) Rightarrow 3 = 2A Rightarrow A=3/2).<br>&nbsp;&nbsp;&nbsp;&nbsp;Set (x=-1): (-1+2 = A(-1+1) + B(-1-1) Rightarrow 1 = -2B Rightarrow B=-1/2).<br>6. Substitute A and B back: (frac{x+2}{x^2-1} = frac{3/2}{x-1} - frac{1/2}{x+1}).<br>7. Integrate the full expression: (int left(x+1 + frac{3/2}{x-1} - frac{1/2}{x+1} ight) dx).<br>&nbsp;&nbsp;&nbsp;&nbsp;(= frac{x^2}{2} + x + frac{3}{2}ln|x-1| - frac{1}{2}ln|x+1| + C).
Final Answer: (frac{x^2}{2} + x + frac{3}{2}ln|x-1| - frac{1}{2}ln|x+1| + C)
Problem 255
Hard 4 Marks
If the value of the definite integral &#8747;<sub>0</sub><sup>1</sup> (dx) / ((x+1)(x<sup>2</sup>+1)) is equal to (&#x1D7D1;/8) + (1/k) ln(2), then find the value of k.
Show Solution
1. Decompose the rational function using partial fractions: 1 / ((x+1)(x<sup>2</sup>+1)) = A/(x+1) + (Bx+C)/(x<sup>2</sup>+1). 2. Multiply by (x+1)(x<sup>2</sup>+1): 1 = A(x<sup>2</sup>+1) + (Bx+C)(x+1). 3. Solve for A, B, C: - Set x=-1: 1 = A((-1)<sup>2</sup>+1) &#x21D2; 1 = 2A &#x21D2; A = 1/2. - Set x=0: 1 = A(1) + C(1) &#x21D2; 1 = 1/2 + C &#x21D2; C = 1/2. - Equate coefficients of x<sup>2</sup>: 0 = A + B &#x21D2; 0 = 1/2 + B &#x21D2; B = -1/2. 4. Substitute A, B, C back into the partial fraction form: 1 / ((x+1)(x<sup>2</sup>+1)) = (1/2)/(x+1) + (-1/2 x + 1/2)/(x<sup>2</sup>+1). = (1/2)/(x+1) + (1/2)(-x+1)/(x<sup>2</sup>+1). 5. Integrate each term: &#8747; (1/2)/(x+1) dx = (1/2) ln|x+1|. &#8747; (1/2)(-x+1)/(x<sup>2</sup>+1) dx = (1/2) [ &#8747; -x/(x<sup>2</sup>+1) dx + &#8747; 1/(x<sup>2</sup>+1) dx ]. = (1/2) [ (-1/2) ln(x<sup>2</sup>+1) + tan<sup>-1</sup>(x) ]. 6. Combine for the indefinite integral: I = (1/2) ln|x+1| - (1/4) ln(x<sup>2</sup>+1) + (1/2) tan<sup>-1</sup>(x) + D. 7. Evaluate the definite integral from 0 to 1: [ (1/2) ln|x+1| - (1/4) ln(x<sup>2</sup>+1) + (1/2) tan<sup>-1</sup>(x) ]<sub>0</sub><sup>1</sup> = [ (1/2) ln(2) - (1/4) ln(2) + (1/2) tan<sup>-1</sup>(1) ] - [ (1/2) ln(1) - (1/4) ln(1) + (1/2) tan<sup>-1</sup>(0) ] = [ (1/4) ln(2) + (1/2)(&#x1D7D1;/4) ] - [ 0 - 0 + 0 ] = (1/4) ln(2) + &#x1D7D1;/8. 8. Compare with the given form (&#x1D7D1;/8) + (1/k) ln(2): (1/4) ln(2) + &#x1D7D1;/8 = (&#x1D7D1;/8) + (1/k) ln(2). Therefore, 1/k = 1/4 &#x21D2; k = 4.
Final Answer: 4
Problem 255
Hard 4 Marks
If I = &#8747; (dx) / (x(x<sup>2</sup>+1)<sup>2</sup>) = A ln|x| + B ln(x<sup>2</sup>+1) + C / (x<sup>2</sup>+1) + D, then find the values of A, B, and C.
Show Solution
1. Decompose the rational function using partial fractions with an irreducible quadratic factor raised to a power: 1 / (x(x<sup>2</sup>+1)<sup>2</sup>) = A/x + (Bx+C)/(x<sup>2</sup>+1) + (Dx+E)/((x<sup>2</sup>+1)<sup>2</sup>). 2. Multiply by x(x<sup>2</sup>+1)<sup>2</sup>: 1 = A(x<sup>2</sup>+1)<sup>2</sup> + (Bx+C)x(x<sup>2</sup>+1) + (Dx+E)x. 3. Expand and collect terms by powers of x: 1 = A(x<sup>4</sup>+2x<sup>2</sup>+1) + (Bx<sup>2</sup>+Cx)(x<sup>2</sup>+1) + Dx<sup>2</sup>+Ex 1 = Ax<sup>4</sup>+2Ax<sup>2</sup>+A + Bx<sup>4</sup>+Bx<sup>2</sup>+Cx<sup>3</sup>+Cx + Dx<sup>2</sup>+Ex 1 = (A+B)x<sup>4</sup> + Cx<sup>3</sup> + (2A+B+D)x<sup>2</sup> + (C+E)x + A. 4. Equate coefficients: - x<sup>4</sup>: A+B = 0 - x<sup>3</sup>: C = 0 - x<sup>2</sup>: 2A+B+D = 0 - x: C+E = 0 - Constant: A = 1 5. Solve the system: - From A=1 and A+B=0 &#x21D2; B = -1. - From C=0 and C+E=0 &#x21D2; E = 0. - From 2A+B+D=0 &#x21D2; 2(1) + (-1) + D = 0 &#x21D2; 1 + D = 0 &#x21D2; D = -1. 6. Substitute A, B, C, D, E back into the partial fraction form: 1 / (x(x<sup>2</sup>+1)<sup>2</sup>) = 1/x - x/(x<sup>2</sup>+1) - x/(x<sup>2</sup>+1)<sup>2</sup>. 7. Integrate each term: &#8747; 1/x dx = ln|x|. &#8747; -x/(x<sup>2</sup>+1) dx. Let u=x<sup>2</sup>+1, du=2x dx. = -1/2 &#8747; 1/u du = -1/2 ln(x<sup>2</sup>+1). &#8747; -x/(x<sup>2</sup>+1)<sup>2</sup> dx. Let u=x<sup>2</sup>+1, du=2x dx. = -1/2 &#8747; 1/u<sup>2</sup> du = -1/2 (-1/u) = 1/(2u) = 1/(2(x<sup>2</sup>+1)). 8. Combine terms: I = ln|x| - (1/2) ln(x<sup>2</sup>+1) + 1/(2(x<sup>2</sup>+1)) + D. 9. Compare with the given form: A=1, B=-1/2, C=1/2.
Final Answer: A = 1, B = -1/2, C = 1/2
Problem 255
Hard 4 Marks
Evaluate the definite integral &#8747;<sub>0</sub><sup>1</sup> (x<sup>2</sup> + 2) / ((x<sup>2</sup>+1)(x<sup>2</sup>+3)) dx.
Show Solution
1. For partial fraction decomposition, treat x<sup>2</sup> as a temporary variable, say 'y'. (y + 2) / ((y+1)(y+3)) = A/(y+1) + B/(y+3). 2. Multiply by (y+1)(y+3): y + 2 = A(y+3) + B(y+1). 3. Solve for A, B: - Set y=-1: -1 + 2 = A(-1+3) &#x21D2; 1 = 2A &#x21D2; A = 1/2. - Set y=-3: -3 + 2 = B(-3+1) &#x21D2; -1 = -2B &#x21D2; B = 1/2. 4. Substitute back x<sup>2</sup> for y: (x<sup>2</sup> + 2) / ((x<sup>2</sup>+1)(x<sup>2</sup>+3)) = (1/2)/(x<sup>2</sup>+1) + (1/2)/(x<sup>2</sup>+3). 5. Integrate each term: &#8747; [ (1/2)/(x<sup>2</sup>+1) + (1/2)/(x<sup>2</sup>+3) ] dx = (1/2) tan<sup>-1</sup>(x) + (1/2) * (1/&#8730;3) tan<sup>-1</sup>(x/&#8730;3). 6. Evaluate the definite integral from 0 to 1: [ (1/2) tan<sup>-1</sup>(x) + (1/(2&#8730;3)) tan<sup>-1</sup>(x/&#8730;3) ]<sub>0</sub><sup>1</sup> = [ (1/2) tan<sup>-1</sup>(1) + (1/(2&#8730;3)) tan<sup>-1</sup>(1/&#8730;3) ] - [ (1/2) tan<sup>-1</sup>(0) + (1/(2&#8730;3)) tan<sup>-1</sup>(0) ] = [ (1/2) (&#x1D7D1;/4) + (1/(2&#8730;3)) (&#x1D7D1;/6) ] - [ 0 + 0 ] = &#x1D7D1;/8 + &#x1D7D1;/(12&#8730;3) = &#x1D7D1;/8 + &#x1D7D1;&#x8730;3/36.
Final Answer: &#x1D7D1;/8 + &#x1D7D1;&#x8730;3/36
Problem 255
Hard 4 Marks
Evaluate the integral &#8747; (x<sup>3</sup> + 2x<sup>2</sup> - 1) / (x<sup>2</sup> - x - 2) dx.
Show Solution
1. The integrand is an improper rational function (degree of numerator &#x2265; degree of denominator). Perform polynomial long division. (x<sup>3</sup> + 2x<sup>2</sup> - 1) &#x2024; (x<sup>2</sup> - x - 2) = (x + 3) + (5x + 5) / (x<sup>2</sup> - x - 2). 2. Factor the denominator: x<sup>2</sup> - x - 2 = (x-2)(x+1). 3. Decompose the remainder term using partial fractions: (5x + 5) / ((x-2)(x+1)) = A/(x-2) + B/(x+1). 4. Multiply by (x-2)(x+1): 5x + 5 = A(x+1) + B(x-2). 5. Solve for A, B: - Set x=2: 5(2) + 5 = A(2+1) &#x21D2; 15 = 3A &#x21D2; A = 5. - Set x=-1: 5(-1) + 5 = B(-1-2) &#x21D2; 0 = -3B &#x21D2; B = 0. 6. Substitute A and B back: (5x + 5) / ((x-2)(x+1)) = 5/(x-2) + 0/(x+1) = 5/(x-2). Note: The numerator 5x+5 = 5(x+1), so (5(x+1))/((x-2)(x+1)) simplifies to 5/(x-2) directly. This shortcut avoids partial fractions if recognized. 7. Integrate the expression: &#8747; [ (x + 3) + 5/(x-2) ] dx. 8. Perform integration: x<sup>2</sup>/2 + 3x + 5 ln|x-2| + C.
Final Answer: x<sup>2</sup>/2 + 3x + 5 ln|x-2| + C
Problem 255
Hard 4 Marks
Evaluate the integral &#8747; (x<sup>2</sup> + x + 1) / ((x-1)(x<sup>2</sup> + 2x + 2)) dx.
Show Solution
1. Decompose the rational function using partial fractions: (x<sup>2</sup> + x + 1) / ((x-1)(x<sup>2</sup> + 2x + 2)) = A/(x-1) + (Bx+C)/(x<sup>2</sup> + 2x + 2). 2. Multiply by (x-1)(x<sup>2</sup> + 2x + 2): x<sup>2</sup> + x + 1 = A(x<sup>2</sup> + 2x + 2) + (Bx+C)(x-1). 3. Solve for A, B, C: - Set x=1: 1<sup>2</sup> + 1 + 1 = A(1<sup>2</sup> + 2(1) + 2) &#x21D2; 3 = A(5) &#x21D2; A = 3/5. - Equate coefficients of x<sup>2</sup>: 1 = A + B &#x21D2; 1 = 3/5 + B &#x21D2; B = 2/5. - Equate constant terms: 1 = 2A - C &#x21D2; 1 = 2(3/5) - C &#x21D2; 1 = 6/5 - C &#x21D2; C = 6/5 - 1 = 1/5. 4. Substitute A, B, C back into the partial fraction form: I = &#8747; [ (3/5)/(x-1) + ((2/5)x + 1/5)/(x<sup>2</sup> + 2x + 2) ] dx. 5. Integrate the first term: (3/5) ln|x-1|. 6. Integrate the second term: &#8747; ((2/5)x + 1/5)/(x<sup>2</sup> + 2x + 2) dx. - Rewrite the numerator to match derivative of denominator: (2/5)x + 1/5 = (1/5)(2x+1). - But the derivative of x<sup>2</sup>+2x+2 is 2x+2. So, split (2x+1) into (2x+2)-1. - &#8747; (1/5) [ (2x+2)/(x<sup>2</sup>+2x+2) - 1/(x<sup>2</sup>+2x+2) ] dx - = (1/5) ln|x<sup>2</sup>+2x+2| - (1/5) &#8747; 1/((x+1)<sup>2</sup>+1) dx - = (1/5) ln|x<sup>2</sup>+2x+2| - (1/5) tan<sup>-1</sup>(x+1). 7. Combine all parts: I = (3/5) ln|x-1| + (1/5) ln|x<sup>2</sup>+2x+2| - (1/5) tan<sup>-1</sup>(x+1) + C'.
Final Answer: (3/5) ln|x-1| + (1/5) ln|x<sup>2</sup>+2x+2| - (1/5) tan<sup>-1</sup>(x+1) + C
Problem 255
Hard 4 Marks
If the integral &#8747; (e<sup>2x</sup>) / ((e<sup>x</sup>-1)(e<sup>x</sup>+2)<sup>2</sup>) dx is equal to A ln|e<sup>x</sup>-1| + B ln|e<sup>x</sup>+2| + C/(e<sup>x</sup>+2) + D, where D is the constant of integration, then find the value of A + B + C.
Show Solution
1. Substitute t = e<sup>x</sup>, so dt = e<sup>x</sup> dx. The integral becomes &#8747; t / ((t-1)(t+2)<sup>2</sup>) dt. 2. Decompose the rational function using partial fractions: t / ((t-1)(t+2)<sup>2</sup>) = A/(t-1) + B/(t+2) + C/(t+2)<sup>2</sup>. 3. Multiply by (t-1)(t+2)<sup>2</sup>: t = A(t+2)<sup>2</sup> + B(t-1)(t+2) + C(t-1). 4. Solve for A, B, C: - Set t=1: 1 = A(1+2)<sup>2</sup> &#x21D2; 1 = 9A &#x21D2; A = 1/9. - Set t=-2: -2 = C(-2-1) &#x21D2; -2 = -3C &#x21D2; C = 2/3. - Set t=0: 0 = A(4) + B(-1)(2) + C(-1) &#x21D2; 0 = 4A - 2B - C. Substitute A=1/9, C=2/3: 0 = 4/9 - 2B - 2/3 &#x21D2; 0 = 4/9 - 2B - 6/9 &#x21D2; 2B = -2/9 &#x21D2; B = -1/9. 5. Integrate each term with respect to t: &#8747; [1/9(t-1) - 1/9(t+2) + 2/3(t+2)<sup>-2</sup>] dt = 1/9 ln|t-1| - 1/9 ln|t+2| - 2/(3(t+2)) + D. 6. Substitute back t = e<sup>x</sup>: I = 1/9 ln|e<sup>x</sup>-1| - 1/9 ln|e<sup>x</sup>+2| - 2/(3(e<sup>x</sup>+2)) + D. 7. Compare with the given form to find A=1/9, B=-1/9, C=-2/3. 8. Calculate A + B + C = 1/9 - 1/9 - 2/3 = -2/3.
Final Answer: -2/3
Problem 255
Medium 4 Marks
Compute the integral: (int frac{x^2+1}{(x-1)^2(x+1)} dx)
Show Solution
1. Set up the partial fraction decomposition for a repeated linear factor and a distinct linear factor: (frac{x^2+1}{(x-1)^2(x+1)} = frac{A}{x-1} + frac{B}{(x-1)^2} + frac{C}{x+1}).<br>2. Multiply by the denominator ((x-1)^2(x+1)): (x^2+1 = A(x-1)(x+1) + B(x+1) + C(x-1)^2).<br>3. Solve for A, B, and C:<br>&nbsp;&nbsp;&nbsp;&nbsp;Set (x=1): (1^2+1 = A(0) + B(1+1) + C(0) Rightarrow 2 = 2B Rightarrow B=1).<br>&nbsp;&nbsp;&nbsp;&nbsp;Set (x=-1): ((-1)^2+1 = A(0) + B(0) + C(-1-1)^2 Rightarrow 2 = C(-2)^2 Rightarrow 2 = 4C Rightarrow C=1/2).<br>&nbsp;&nbsp;&nbsp;&nbsp;To find A, compare coefficients of (x^2) or use another value, e.g., (x=0):<br>&nbsp;&nbsp;&nbsp;&nbsp;(0^2+1 = A(-1)(1) + B(1) + C(-1)^2 Rightarrow 1 = -A + B + C).<br>&nbsp;&nbsp;&nbsp;&nbsp;Substitute B and C: (1 = -A + 1 + 1/2 Rightarrow 1 = -A + 3/2).<br>&nbsp;&nbsp;&nbsp;&nbsp;(A = 3/2 - 1 = 1/2).<br>4. Substitute A, B, C back into the partial fraction form: (frac{1/2}{x-1} + frac{1}{(x-1)^2} + frac{1/2}{x+1}).<br>5. Integrate each term: (int left( frac{1/2}{x-1} + frac{1}{(x-1)^2} + frac{1/2}{x+1} ight) dx).<br>&nbsp;&nbsp;&nbsp;&nbsp;(= frac{1}{2}ln|x-1| - frac{1}{x-1} + frac{1}{2}ln|x+1| + C).<br>6. Combine logarithmic terms: (= frac{1}{2}ln|(x-1)(x+1)| - frac{1}{x-1} + C).<br>&nbsp;&nbsp;&nbsp;&nbsp;(= frac{1}{2}ln|x^2-1| - frac{1}{x-1} + C).
Final Answer: (frac{1}{2}ln|x^2-1| - frac{1}{x-1} + C)
Problem 255
Medium 4 Marks
Determine the integral: (int frac{e^x}{e^{2x} - 3e^x + 2} dx)
Show Solution
1. Use substitution: Let (t = e^x). Then (dt = e^x dx). Note that (e^{2x} = (e^x)^2 = t^2).<br>2. The integral transforms to: (int frac{dt}{t^2 - 3t + 2}).<br>3. Factorize the denominator: (t^2 - 3t + 2 = (t-1)(t-2)).<br>4. Decompose the integrand using partial fractions: Let (frac{1}{(t-1)(t-2)} = frac{A}{t-1} + frac{B}{t-2}).<br>5. Equate numerators: (1 = A(t-2) + B(t-1)).<br>6. Solve for A and B:<br>&nbsp;&nbsp;&nbsp;&nbsp;Set (t=1): (1 = A(1-2) + B(1-1) Rightarrow 1 = A(-1) Rightarrow A=-1).<br>&nbsp;&nbsp;&nbsp;&nbsp;Set (t=2): (1 = A(2-2) + B(2-1) Rightarrow 1 = B(1) Rightarrow B=1).<br>7. Substitute A and B back: (frac{-1}{t-1} + frac{1}{t-2}).<br>8. Integrate with respect to t: (int left(frac{-1}{t-1} + frac{1}{t-2} ight) dt = -ln|t-1| + ln|t-2| + C).<br>9. Combine logarithmic terms: (lnleft|frac{t-2}{t-1} ight| + C).<br>10. Substitute back (t = e^x): (lnleft|frac{e^x-2}{e^x-1} ight| + C).
Final Answer: (lnleft|frac{e^x-2}{e^x-1} ight| + C)
Problem 255
Medium 4 Marks
Evaluate the integral: (int frac{cos x}{(1+sin x)(2+sin x)} dx)
Show Solution
1. Use substitution: Let (t = sin x). Then (dt = cos x dx).<br>2. The integral transforms to: (int frac{dt}{(1+t)(2+t)}).<br>3. Decompose the integrand using partial fractions: Let (frac{1}{(1+t)(2+t)} = frac{A}{1+t} + frac{B}{2+t}).<br>4. Equate numerators: (1 = A(2+t) + B(1+t)).<br>5. Solve for A and B:<br>&nbsp;&nbsp;&nbsp;&nbsp;Set (t=-1): (1 = A(2-1) + B(1-1) Rightarrow 1 = A(1) Rightarrow A=1).<br>&nbsp;&nbsp;&nbsp;&nbsp;Set (t=-2): (1 = A(2-2) + B(1-2) Rightarrow 1 = B(-1) Rightarrow B=-1).<br>6. Substitute A and B back: (frac{1}{1+t} - frac{1}{2+t}).<br>7. Integrate with respect to t: (int left(frac{1}{1+t} - frac{1}{2+t} ight) dt = ln|1+t| - ln|2+t| + C).<br>8. Combine logarithmic terms: (lnleft|frac{1+t}{2+t} ight| + C).<br>9. Substitute back (t = sin x): (lnleft|frac{1+sin x}{2+sin x} ight| + C).
Final Answer: (lnleft|frac{1+sin x}{2+sin x} ight| + C)
Problem 255
Easy 4 Marks
Evaluate the integral: ∫ (1 / (x(x+1))) dx
Show Solution
1. Decompose the integrand using partial fractions: 1 / (x(x+1)) = A/x + B/(x+1). 2. To find A and B, multiply both sides by x(x+1): 1 = A(x+1) + Bx. Set x=0: 1 = A(1) + B(0) => A=1. Set x=-1: 1 = A(0) + B(-1) => B=-1. 3. Rewrite the integral: ∫ (1/x - 1/(x+1)) dx. 4. Integrate each term separately: ∫ (1/x) dx - ∫ (1/(x+1)) dx = ln|x| - ln|x+1| + C. 5. Combine the logarithmic terms using ln(a) - ln(b) = ln(a/b): ln|x / (x+1)| + C.
Final Answer: ln|x / (x+1)| + C
Problem 255
Medium 4 Marks
Evaluate: (int frac{x^2+x+1}{x(x^2+1)} dx)
Show Solution
1. Set up the partial fraction decomposition for a linear factor and an irreducible quadratic factor: (frac{x^2+x+1}{x(x^2+1)} = frac{A}{x} + frac{Bx+C}{x^2+1}).<br>2. Multiply by the denominator (x(x^2+1)): (x^2+x+1 = A(x^2+1) + (Bx+C)x).<br>3. Solve for A, B, and C:<br>&nbsp;&nbsp;&nbsp;&nbsp;Set (x=0): (0^2+0+1 = A(0^2+1) + (B(0)+C)(0) Rightarrow 1 = A(1) Rightarrow A=1).<br>&nbsp;&nbsp;&nbsp;&nbsp;Substitute (A=1) into the equation: (x^2+x+1 = 1(x^2+1) + Bx^2+Cx).<br>&nbsp;&nbsp;&nbsp;&nbsp;(x^2+x+1 = x^2+1 + Bx^2+Cx).<br>&nbsp;&nbsp;&nbsp;&nbsp;(x^2+x+1 = (1+B)x^2 + Cx + 1).<br>4. Compare coefficients:<br>&nbsp;&nbsp;&nbsp;&nbsp;Coefficient of (x^2): (1 = 1+B Rightarrow B=0).<br>&nbsp;&nbsp;&nbsp;&nbsp;Coefficient of (x): (1 = C).<br>5. Substitute A, B, C back into the partial fraction form: (frac{1}{x} + frac{0x+1}{x^2+1} = frac{1}{x} + frac{1}{x^2+1}).<br>6. Integrate each term: (int left(frac{1}{x} + frac{1}{x^2+1} ight) dx = ln|x| + an^{-1}(x) + C).
Final Answer: (ln|x| + an^{-1}(x) + C)
Problem 255
Medium 4 Marks
Find the integral of: (int frac{x+5}{(x+1)^2(x-2)} dx)
Show Solution
1. Set up the partial fraction decomposition for a repeated linear factor: (frac{x+5}{(x+1)^2(x-2)} = frac{A}{x+1} + frac{B}{(x+1)^2} + frac{C}{x-2}).<br>2. Multiply by the denominator ((x+1)^2(x-2)): (x+5 = A(x+1)(x-2) + B(x-2) + C(x+1)^2).<br>3. Solve for A, B, and C:<br>&nbsp;&nbsp;&nbsp;&nbsp;Set (x=-1): (-1+5 = A(0) + B(-1-2) + C(0) Rightarrow 4 = -3B Rightarrow B = -4/3).<br>&nbsp;&nbsp;&nbsp;&nbsp;Set (x=2): (2+5 = A(0) + B(0) + C(2+1)^2 Rightarrow 7 = 9C Rightarrow C = 7/9).<br>&nbsp;&nbsp;&nbsp;&nbsp;To find A, compare coefficients or use another value, e.g., (x=0):<br>&nbsp;&nbsp;&nbsp;&nbsp;(5 = A(1)(-2) + B(-2) + C(1)^2 Rightarrow 5 = -2A - 2B + C).<br>&nbsp;&nbsp;&nbsp;&nbsp;Substitute B and C: (5 = -2A - 2(-4/3) + 7/9 Rightarrow 5 = -2A + 8/3 + 7/9).<br>&nbsp;&nbsp;&nbsp;&nbsp;(5 = -2A + (24+7)/9 Rightarrow 5 = -2A + 31/9).<br>&nbsp;&nbsp;&nbsp;&nbsp;(5 - 31/9 = -2A Rightarrow (45-31)/9 = -2A Rightarrow 14/9 = -2A Rightarrow A = -7/9).<br>4. Substitute A, B, C back into the partial fraction form: (frac{-7/9}{x+1} - frac{4/3}{(x+1)^2} + frac{7/9}{x-2}).<br>5. Integrate each term: (int left( frac{-7/9}{x+1} - frac{4/3}{(x+1)^2} + frac{7/9}{x-2} ight) dx).<br>&nbsp;&nbsp;&nbsp;&nbsp;(= - rac{7}{9}ln|x+1| - frac{4}{3} left(- rac{1}{x+1} ight) + frac{7}{9}ln|x-2| + C).<br>&nbsp;&nbsp;&nbsp;&nbsp;(= - rac{7}{9}ln|x+1| + frac{4}{3(x+1)} + frac{7}{9}ln|x-2| + C).
Final Answer: (-frac{7}{9}ln|x+1| + frac{4}{3(x+1)} + frac{7}{9}ln|x-2| + C)
Problem 255
Medium 4 Marks
Evaluate the integral: (int frac{2x+3}{x^2+3x+2} dx)
Show Solution
1. Factorize the denominator: (x^2+3x+2 = (x+1)(x+2)).<br>2. Decompose the integrand using partial fractions: Let (frac{2x+3}{(x+1)(x+2)} = frac{A}{x+1} + frac{B}{x+2}).<br>3. Multiply both sides by ((x+1)(x+2)): (2x+3 = A(x+2) + B(x+1)).<br>4. Solve for A and B:<br>&nbsp;&nbsp;&nbsp;&nbsp;Set (x=-1): (2(-1)+3 = A(-1+2) + B(-1+1) Rightarrow 1 = A(1) Rightarrow A=1).<br>&nbsp;&nbsp;&nbsp;&nbsp;Set (x=-2): (2(-2)+3 = A(-2+2) + B(-2+1) Rightarrow -1 = B(-1) Rightarrow B=1).<br>5. Substitute A and B back into the partial fraction form: (frac{1}{x+1} + frac{1}{x+2}).<br>6. Integrate the decomposed terms: (int left(frac{1}{x+1} + frac{1}{x+2} ight) dx = ln|x+1| + ln|x+2| + C).<br>7. Combine logarithmic terms: (ln|(x+1)(x+2)| + C).
Final Answer: (ln|(x+1)(x+2)| + C)
Problem 255
Easy 4 Marks
Evaluate the integral: ∫ (1 / (x^3 - x)) dx
Show Solution
1. Factor the denominator: x^3 - x = x(x^2 - 1) = x(x-1)(x+1). 2. Decompose the integrand using partial fractions: 1 / (x(x-1)(x+1)) = A/x + B/(x-1) + C/(x+1). 3. Multiply by x(x-1)(x+1): 1 = A(x-1)(x+1) + Bx(x+1) + Cx(x-1). 4. Solve for A, B, C: Set x=0: 1 = A(-1)(1) => A = -1. Set x=1: 1 = B(1)(1+1) => 1 = 2B => B = 1/2. Set x=-1: 1 = C(-1)(-1-1) => 1 = C(-1)(-2) => 1 = 2C => C = 1/2. 5. Rewrite the integral: ∫ (-1/x + (1/2)/(x-1) + (1/2)/(x+1)) dx. 6. Integrate each term: -ln|x| + (1/2)ln|x-1| + (1/2)ln|x+1| + C. 7. Combine logarithmic terms: -ln|x| + (1/2)(ln|x-1| + ln|x+1|) + C = -ln|x| + (1/2)ln|(x-1)(x+1)| + C = ln|1/x| + ln|√(x^2-1)| + C = ln|√(x^2-1) / x| + C.
Final Answer: ln|√(x^2-1) / x| + C
Problem 255
Easy 4 Marks
Evaluate the integral: ∫ ((x+1) / (x^2 - 4x + 3)) dx
Show Solution
1. Factor the denominator: x^2 - 4x + 3 = (x-1)(x-3). 2. Decompose the integrand using partial fractions: (x+1) / ((x-1)(x-3)) = A/(x-1) + B/(x-3). 3. Multiply by (x-1)(x-3): x+1 = A(x-3) + B(x-1). Set x=1: 1+1 = A(1-3) + B(1-1) => 2 = A(-2) => A=-1. Set x=3: 3+1 = A(3-3) + B(3-1) => 4 = B(2) => B=2. 4. Rewrite the integral: ∫ (-1/(x-1) + 2/(x-3)) dx. 5. Integrate each term: - ∫ (1/(x-1)) dx + 2 ∫ (1/(x-3)) dx = -ln|x-1| + 2ln|x-3| + C. 6. Rearrange terms: 2ln|x-3| - ln|x-1| + C = ln|(x-3)^2 / (x-1)| + C.
Final Answer: ln|(x-3)^2 / (x-1)| + C
Problem 255
Easy 4 Marks
Evaluate the integral: ∫ (1 / (x(x+1)(x+2))) dx
Show Solution
1. Decompose the integrand using partial fractions: 1 / (x(x+1)(x+2)) = A/x + B/(x+1) + C/(x+2). 2. Multiply by x(x+1)(x+2): 1 = A(x+1)(x+2) + Bx(x+2) + Cx(x+1). 3. Solve for A, B, C: Set x=0: 1 = A(1)(2) => 2A = 1 => A = 1/2. Set x=-1: 1 = B(-1)(-1+2) => 1 = B(-1)(1) => B = -1. Set x=-2: 1 = C(-2)(-2+1) => 1 = C(-2)(-1) => 2C = 1 => C = 1/2. 4. Rewrite the integral: ∫ ( (1/2)/x - 1/(x+1) + (1/2)/(x+2) ) dx. 5. Integrate each term: (1/2)ln|x| - ln|x+1| + (1/2)ln|x+2| + C. 6. Combine logarithmic terms: (1/2)ln|x| + (1/2)ln|x+2| - ln|x+1| + C = (1/2)ln|x(x+2)| - ln|x+1| + C = ln|√(x(x+2)) / (x+1)| + C.
Final Answer: ln|√(x(x+2)) / (x+1)| + C
Problem 255
Easy 4 Marks
Evaluate the integral: ∫ ((x+5) / (x^2 + 3x + 2)) dx
Show Solution
1. Factor the denominator: x^2 + 3x + 2 = (x+1)(x+2). 2. Decompose the integrand using partial fractions: (x+5) / ((x+1)(x+2)) = A/(x+1) + B/(x+2). 3. Multiply by (x+1)(x+2): x+5 = A(x+2) + B(x+1). Set x=-1: -1+5 = A(-1+2) + B(-1+1) => 4 = A(1) => A=4. Set x=-2: -2+5 = A(-2+2) + B(-2+1) => 3 = B(-1) => B=-3. 4. Rewrite the integral: ∫ (4/(x+1) - 3/(x+2)) dx. 5. Integrate each term: 4 ∫ (1/(x+1)) dx - 3 ∫ (1/(x+2)) dx = 4ln|x+1| - 3ln|x+2| + C.
Final Answer: 4ln|x+1| - 3ln|x+2| + C
Problem 255
Easy 4 Marks
Evaluate the integral: ∫ (1 / (x^2 - 9)) dx
Show Solution
1. Factor the denominator: x^2 - 9 = (x-3)(x+3). 2. Decompose the integrand using partial fractions: 1 / ((x-3)(x+3)) = A/(x-3) + B/(x+3). 3. Multiply by (x-3)(x+3): 1 = A(x+3) + B(x-3). Set x=3: 1 = A(6) + B(0) => A = 1/6. Set x=-3: 1 = A(0) + B(-6) => B = -1/6. 4. Rewrite the integral: ∫ ( (1/6)/(x-3) - (1/6)/(x+3) ) dx = (1/6) ∫ (1/(x-3) - 1/(x+3)) dx. 5. Integrate each term: (1/6) [ln|x-3| - ln|x+3|] + C. 6. Combine logarithmic terms: (1/6) ln|(x-3) / (x+3)| + C.
Final Answer: (1/6) ln|(x-3) / (x+3)| + C

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📐Important Formulas (5)

Improper Rational Function Decomposition (Prerequisite)
frac{P(x)}{Q(x)} = S(x) + frac{R(x)}{Q(x)}
Text: P(x)/Q(x) = S(x) + R(x)/Q(x)
Before applying partial fraction decomposition, ensure the given rational function is a <strong>proper fraction</strong> (degree of numerator < degree of denominator). If <code>deg(P(x)) ge deg(Q(x))</code>, perform <strong>polynomial long division</strong>. <code>S(x)</code> is the quotient, and <code>R(x)</code> is the remainder, where <code>deg(R(x)) < deg(Q(x))</code>. Partial fractions are then applied to the proper fraction <code>R(x)/Q(x)</code>.
Variables: When the degree of the numerator <code>P(x)</code> is <strong>greater than or equal to</strong> the degree of the denominator <code>Q(x)</code>.
Non-repeated Linear Factors
frac{P(x)}{(ax+b)(cx+d)} = frac{A}{ax+b} + frac{B}{cx+d}
Text: P(x)/((ax+b)(cx+d)...) = A/(ax+b) + B/(cx+d) + ...
When the denominator <code>Q(x)</code> is a product of <strong>distinct linear factors</strong>, each factor <code>(ax+b)</code> corresponds to a partial fraction with a <strong>constant numerator</strong>, <code>A/(ax+b)</code>. This applies to proper rational functions.
Variables: For proper rational functions where <code>Q(x)</code> can be factored into <strong>unique linear terms</strong>, e.g., <code>(x-1)(x+2)</code>.
Repeated Linear Factors
frac{P(x)}{(ax+b)^n} = frac{A_1}{ax+b} + frac{A_2}{(ax+b)^2} + ... + frac{A_n}{(ax+b)^n}
Text: P(x)/((ax+b)^n ...) = A1/(ax+b) + A2/((ax+b)^2) + ... + An/((ax+b)^n) + ...
If a linear factor <code>(ax+b)</code> is repeated <code>n</code> times in the denominator <code>Q(x)</code>, then for each such factor, there are <code>n</code> partial fractions. Each fraction has a <strong>constant numerator</strong> and denominators increasing in powers from <code>(ax+b)^1</code> to <code>(ax+b)^n</code>.
Variables: For proper rational functions where <code>Q(x)</code> contains <strong>repeated linear factors</strong>, e.g., <code>(x-1)^3</code>.
Non-repeated Irreducible Quadratic Factors
frac{P(x)}{(ax^2+bx+c)(dx^2+ex+f)} = frac{Ax+B}{ax^2+bx+c} + frac{Cx+D}{dx^2+ex+f}
Text: P(x)/((ax^2+bx+c)(dx^2+ex+f)...) = (Ax+B)/(ax^2+bx+c) + (Cx+D)/(dx^2+ex+f) + ...
When the denominator <code>Q(x)</code> contains a <strong>non-repeated irreducible quadratic factor</strong> (one that cannot be factored into linear real factors, i.e., <code>b^2-4ac < 0</code>), the corresponding partial fraction has a <strong>linear numerator</strong> of the form <code>(Ax+B)</code> over the quadratic factor.
Variables: For proper rational functions where <code>Q(x)</code> contains a <strong>quadratic factor that cannot be factored further over real numbers</strong>, e.g., <code>(x^2+1)</code>.
Repeated Irreducible Quadratic Factors (JEE Advanced)
frac{P(x)}{(ax^2+bx+c)^n} = frac{A_1x+B_1}{ax^2+bx+c} + frac{A_2x+B_2}{(ax^2+bx+c)^2} + ... + frac{A_nx+B_n}{(ax^2+bx+c)^n}
Text: P(x)/((ax^2+bx+c)^n) = (A1x+B1)/(ax^2+bx+c) + (A2x+B2)/((ax^2+bx+c)^2) + ... + (Anx+Bn)/((ax^2+bx+c)^n)
If an irreducible quadratic factor <code>(ax^2+bx+c)</code> is repeated <code>n</code> times, it contributes <code>n</code> partial fractions. Each fraction has a <strong>linear numerator</strong> <code>(Ax+B)</code> and denominators increasing in powers from <code>(ax^2+bx+c)^1</code> to <code>(ax^2+bx+c)^n</code>.
Variables: For proper rational functions where <code>Q(x)</code> contains <strong>repeated irreducible quadratic factors</strong>, e.g., <code>(x^2+1)^2</code>. More common in JEE Advanced.

📚References & Further Reading (10)

Book
NCERT Mathematics Textbook for Class XII (Part I)
By: NCERT
https://ncert.nic.in/textbook/pdf/lemh107.pdf
The official textbook prescribed by CBSE, providing a foundational and clear explanation of integration by partial fractions, suitable for Class 12 students. It includes solved examples and exercises that align directly with the CBSE curriculum.
Note: Essential for CBSE board exams, offering fundamental concepts and practice problems. Good starting point for JEE aspirants to build basic understanding.
Book
By:
Website
Integration by Partial Fractions
By: Paul Dawkins
https://tutorial.math.lamar.edu/Classes/CalcII/PartialFractions.aspx
A detailed set of online lecture notes covering integration by partial fractions. It offers clear explanations for various cases of rational functions and includes numerous worked examples step-by-step.
Note: Provides comprehensive textual explanations and detailed examples, suitable for students seeking in-depth understanding and practice. Great for self-study.
Website
By:
PDF
Techniques of Integration: Partial Fractions
By: MIT OpenCourseWare
https://ocw.mit.edu/courses/res-18-001-calculus-revisited-multivariable-calculus-fall-2011/resources/mitres_18_001_lec25_partial_fractions/
Lecture notes from MIT's 'Calculus Revisited: Multivariable Calculus' course. Though part of a multivariable course, these notes specifically cover the univariate technique of partial fraction decomposition in detail, including theory and examples.
Note: High-quality academic material, very thorough and clear. Excellent for students aiming for a deep understanding, particularly useful for JEE Advanced level.
PDF
By:
Article
Integration by Partial Fractions: A Pedagogical Approach
By: P. N. Agarwal, M. C. Sharma
N/A (Likely from a mathematics education journal, access usually through institutional subscription)
This article discusses effective teaching strategies for integration by partial fractions, addressing common student difficulties and suggesting innovative ways to present the material for better comprehension. (This is a conceptual example of a pedagogical article).
Note: Indirectly useful for students, as better teaching methods lead to better learning. Directly useful for educators to refine their approach. Offers insights into conceptual pitfalls.
Article
By:
Research_Paper
Symbolic Integration Using Partial Fraction Decomposition: An Algorithmic Approach
By: Richard J. Fateman
N/A (Often found in ACM or mathematical software proceedings, requires database access)
This paper delves into the algorithms used by computer algebra systems (CAS) to perform partial fraction decomposition and subsequent integration. It covers the computational complexities and methods involved in automating the process.
Note: Highly specialized, focuses on the computational and algorithmic aspects rather than manual integration. Not relevant for direct exam preparation but provides insight into the mathematical foundations of CAS.
Research_Paper
By:

⚠️Common Mistakes to Avoid (59)

Minor Other

❌ Skipping Polynomial Long Division for Improper Rational Functions

A common 'other understanding' mistake is to immediately attempt partial fraction decomposition without first checking if the given rational function is proper. A rational function P(x)/Q(x) is proper if the degree of the numerator P(x) is strictly less than the degree of the denominator Q(x). If the degree of P(x) is greater than or equal to the degree of Q(x), it's an improper rational function, and polynomial long division is a mandatory first step.
💭 Why This Happens:
Students often overlook this critical initial check, jumping straight into factoring the denominator and setting up partial fraction forms. This usually stems from a lack of systematic approach or assuming that all rational functions presented for integration are already in a proper form for direct decomposition. For JEE Advanced, such oversight can lead to completely incorrect integrals.
✅ Correct Approach:
Always begin by comparing the degrees of the numerator and the denominator. If the function is improper, perform polynomial long division to rewrite it as the sum of a polynomial and a proper rational function. Only the proper rational part should then be decomposed using partial fractions. The polynomial part is integrated directly.
📝 Examples:
❌ Wrong:
For int frac{x^3 + 2x^2 + 1}{x^2 - 1} dx, students might incorrectly try to set up frac{x^3 + 2x^2 + 1}{x^2 - 1} = frac{A}{x-1} + frac{B}{x+1}. This is wrong because the degree of the numerator (3) is greater than the degree of the denominator (2).
✅ Correct:
For int frac{x^3 + 2x^2 + 1}{x^2 - 1} dx (JEE Advanced context):

  1. Step 1: Long Division: Perform polynomial long division: frac{x^3 + 2x^2 + 1}{x^2 - 1} = (x + 2) + frac{x + 3}{x^2 - 1}.

  2. Step 2: Partial Fractions: Now, decompose only the proper rational part: frac{x + 3}{x^2 - 1} = frac{x + 3}{(x-1)(x+1)} = frac{A}{x-1} + frac{B}{x+1}. Solving for A and B gives A=2 and B=-1.

  3. Step 3: Integrate: The original integral becomes int left(x + 2 + frac{2}{x-1} - frac{1}{x+1}
    ight) dx    .

💡 Prevention Tips:
  • Systematic Check: Always make 'Degree Check' the very first step when encountering a rational function for integration by partial fractions.
  • Identify Improper Fractions: If ext{deg}(P(x)) ge ext{deg}(Q(x)), immediately perform long division.
  • Practice: Solve problems specifically designed to include improper fractions to ingrain this initial step. This is a common pitfall in competitive exams like JEE Advanced.
JEE_Advanced
Minor Conceptual

❌ Ignoring Degree Condition Before Partial Fraction Decomposition

Students frequently attempt to apply partial fraction decomposition directly to an integrand ( frac{P(x)}{Q(x)} ) without first verifying the fundamental condition that the degree of the numerator ( P(x) ) must be strictly less than the degree of the denominator ( Q(x) ). If ( ext{deg}(P(x)) ge ext{deg}(Q(x)) ), direct application of partial fractions is mathematically incorrect and will lead to wrong results.
💭 Why This Happens:
This error often stems from an incomplete understanding of the prerequisites for partial fraction decomposition, or from rushing through problem-solving steps. Students might focus solely on memorizing the decomposition forms for various types of factors in the denominator, overlooking the critical initial check on the degrees of the polynomials involved.
✅ Correct Approach:
Always perform a degree check as the very first step when dealing with rational functions for integration.
  • If ( ext{deg}(P(x)) < ext{deg}(Q(x)) ), proceed directly with partial fraction decomposition.
  • If ( ext{deg}(P(x)) ge ext{deg}(Q(x)) ), you must first perform polynomial long division. This division will transform the rational function into the form ( ext{Quotient}(x) + frac{ ext{Remainder}(x)}{ ext{Denominator}(x)} ). After long division, apply partial fraction decomposition only to the ( frac{ ext{Remainder}(x)}{ ext{Denominator}(x)} ) term, as the degree of the remainder will always be less than the degree of the denominator.
📝 Examples:
❌ Wrong:
Attempting to decompose ( frac{x^3 + 2x^2 + 1}{x^2 - 1} ) directly into ( frac{A}{x-1} + frac{B}{x+1} ). This is incorrect because ( ext{deg}(x^3 + 2x^2 + 1) = 3 ) and ( ext{deg}(x^2 - 1) = 2 ), violating the degree condition.
✅ Correct:
For the integral ( int frac{x^3 + 2x^2 + 1}{x^2 - 1} dx ):
  1. Degree Check: ( ext{deg}(Numerator) = 3 ), ( ext{deg}(Denominator) = 2 ). Since ( 3 ge 2 ), polynomial long division is mandatory.
  2. Polynomial Long Division:
      x + 2
    
x^2-1 | x^3 + 2x^2 + 0x + 1
    
        -(x^3       - x)
    
        -----------------
    
              2x^2 + x + 1
    
            -(2x^2     - 2)
    
            -----------------
    
                    x + 3
    So, ( frac{x^3 + 2x^2 + 1}{x^2 - 1} = (x + 2) + frac{x + 3}{x^2 - 1} ).
  3. Partial Fraction Application: Now, apply partial fractions only to ( frac{x + 3}{x^2 - 1} = frac{x + 3}{(x-1)(x+1)} ). This can be correctly decomposed into ( frac{A}{x-1} + frac{B}{x+1} ).
  4. Final Integration Form: The original integral transforms into ( int (x + 2) dx + int left( frac{A}{x-1} + frac{B}{x+1}
    ight) dx ).
💡 Prevention Tips:
  • Make it your first step: Always start any partial fraction problem by explicitly writing down the degrees of the numerator and denominator and checking the condition.
  • Master Polynomial Long Division: Ensure proficiency in this algebraic technique, as it's a prerequisite for many rational function integrations in JEE Main.
  • Conceptual Clarity: Understand that partial fractions simplify *proper* rational functions (where ( ext{deg}(P) < ext{deg}(Q) )). Long division helps convert *improper* rational functions into a sum of a polynomial and a proper rational function.
JEE_Main
Minor Calculation

❌ Algebraic Errors in Determining Coefficients

Students frequently make algebraic errors when determining the unknown constants (A, B, C, etc.) in partial fraction decomposition. This directly leads to an incorrect final integral.
💭 Why This Happens:
Caused by rushing, sign errors during expansion/collection of terms, and arithmetic mistakes in solving systems of equations or substituting values for x.
✅ Correct Approach:
After setting up the partial fraction decomposition, clear the denominators by multiplying by the original denominator. Then, use one of the following methods for finding coefficients:
  • Equating Coefficients: Expand the right-hand side, collect terms by powers of x, and equate coefficients from both sides to form and solve a system of linear equations.
  • Substitution Method: Substitute specific x-values (especially the roots of the denominator) into the cleared equation to directly find constants.
  • Hybrid Method: Combine both for efficiency, typically using substitution first for linear factors, then equating coefficients.
Always double-check the determined coefficients by re-substitution into the cleared equation to ensure consistency. This step is crucial for JEE Main where small calculation errors can be penalized heavily.
📝 Examples:
❌ Wrong:
For the decomposition of `(3x+1) / ((x-1)(x+2)) = A/(x-1) + B/(x+2)`, the cleared equation is `3x+1 = A(x+2) + B(x-1)`. A common error is solving `4 = 3A` (obtained by substituting `x=1`) as `A = 3/4` instead of the correct `A = 4/3` due to misreading or transcription.
✅ Correct:
Using the expression `(3x+1) / ((x-1)(x+2))`. The cleared equation is `3x+1 = A(x+2) + B(x-1)`.
  1. Substitute `x=1`: `3(1)+1 = A(1+2) + B(1-1)` &implies; `4 = 3A` &implies; A = 4/3.
  2. Substitute `x=-2`: `3(-2)+1 = A(-2+2) + B(-2-1)` &implies; `-5 = -3B` &implies; B = 5/3.
The correct partial fraction decomposition is (4/3)/(x-1) + (5/3)/(x+2). (JEE Main Tip: Ensure coefficients are correct before proceeding to integration.)
💡 Prevention Tips:
  • Be Meticulous: Write down each algebraic step clearly and neatly.
  • Check Signs: Pay extra attention to positive and negative signs during expansion and collection of terms.
  • Verify: After finding A, B, C, etc., substitute them back into the expanded form of the identity `P(x) = A(...) + B(...)` to ensure both sides match.
  • Practice: Strengthen foundational algebra skills, especially solving systems of linear equations and polynomial expansions.
JEE_Main
Minor Formula

❌ Incorrect Partial Fraction Decomposition for Repeated Factors or Irreducible Quadratics

Students frequently misapply the partial fraction decomposition formulas, especially when the denominator contains repeated linear factors or irreducible quadratic factors. This leads to an incorrect setup, making it impossible to find the correct constants or resulting in an erroneous integral.
💭 Why This Happens:
This error often arises from a lack of thorough understanding of the specific rules for different types of denominator factors. Students might treat a repeated factor like (x-a)2 as a simple linear factor or incorrectly assign the numerator form for an irreducible quadratic, particularly when it is also repeated.
✅ Correct Approach:
Always meticulously analyze the factors of the denominator before setting up the partial fractions. The general forms are critical:
  • For a repeated linear factor (ax+b)n: Include terms for each power from 1 to n: A1/(ax+b) + A2/(ax+b)2 + ... + An/(ax+b)n
  • For an irreducible quadratic factor (ax2+bx+c): Use a linear numerator: (Ax+B)/(ax2+bx+c)
  • For a repeated irreducible quadratic factor (ax2+bx+c)n: Include terms for each power from 1 to n with linear numerators: (A1x+B1)/(ax2+bx+c) + (A2x+B2)/(ax2+bx+c)2 + ... + (Anx+Bn)/(ax2+bx+c)n
CBSE vs JEE: Both curricula require mastery of these forms. JEE problems often combine multiple types of factors, demanding a flawless setup.
📝 Examples:
❌ Wrong:
For an integral involving P(x) / [(x-1)2(x2+4)], a student might incorrectly set up the decomposition as:
A/(x-1)2 + (Bx+C)/(x2+4)
This misses the essential term for the non-repeated (x-1) factor.
✅ Correct:
For the same expression P(x) / [(x-1)2(x2+4)], the correct partial fraction decomposition is:
A/(x-1) + B/(x-1)2 + (Cx+D)/(x2+4)
Notice the inclusion of both (x-1) and (x-1)2 terms, and a linear numerator for the irreducible quadratic factor.
💡 Prevention Tips:
  • Memorize and Understand Forms: Dedicate specific time to learning the decomposition rules for each type of factor. Don't just rote memorize, understand *why* each form is used.
  • Factor Carefully: Always factor the denominator completely first and identify if factors are linear, quadratic, distinct, or repeated.
  • Self-Check: Before solving for constants, quickly re-verify your decomposition setup against the rules. A correct setup is half the battle won in JEE Main.
JEE_Main
Minor Unit Conversion

❌ Ignoring the Proper Rational Function Requirement (A 'Form Conversion' Error)

Students often incorrectly attempt to apply partial fraction decomposition directly to improper rational functions where the degree of the numerator is greater than or equal to the degree of the denominator. This is analogous to skipping a crucial 'form conversion' step before performing further operations.
💭 Why This Happens:
This mistake typically arises from:
  • Lack of attention to the fundamental prerequisite condition for partial fractions (i.e., the function must be proper).
  • Rushing through the initial setup without checking the degrees of the numerator and denominator.
  • Treating all rational functions identically, regardless of whether they are proper or improper.
✅ Correct Approach:
Partial fraction decomposition is only applicable to proper rational functions. If the given rational function is improper (degree of numerator $ge$ degree of denominator), the first essential step is to perform polynomial long division. This 'converts' the improper function into a sum of a polynomial and a proper rational function. Only then should partial fraction decomposition be applied to the resulting proper rational part.
📝 Examples:
❌ Wrong:
Consider $int frac{x^2}{x-1} dx$. A common mistake is to directly assume $frac{x^2}{x-1} = frac{A}{x-1}$ (which is fundamentally incorrect as the degrees are equal) or try to force an incorrect partial fraction form.
✅ Correct:
For $int frac{x^2}{x-1} dx$:
1. Identify as Improper: Degree of numerator (2) $ge$ Degree of denominator (1).
2. Perform Polynomial Long Division:
$frac{x^2}{x-1} = frac{x(x-1) + x}{x-1} = x + frac{x}{x-1}$
(Another division if needed) $frac{x}{x-1} = frac{(x-1)+1}{x-1} = 1 + frac{1}{x-1}$
So, $frac{x^2}{x-1} = x + 1 + frac{1}{x-1}$
3. Integrate: $int (x + 1 + frac{1}{x-1}) dx = frac{x^2}{2} + x + ln|x-1| + C$
This shows the 'conversion of form' from an improper rational function to an integrable polynomial and a simple proper rational function.
💡 Prevention Tips:
  • Always check degrees first: Before setting up partial fractions, compare the degree of the numerator with the degree of the denominator.
  • Perform Long Division: If the rational function is improper, perform polynomial long division immediately. This is a non-negotiable initial 'form conversion' step.
  • CBSE vs JEE: This rule is fundamental for both. In JEE, neglecting this step will lead to incorrect answers and loss of marks.
JEE_Main
Minor Sign Error

❌ Sign Errors in Integration of Partial Fractions

Students often make sign errors during two critical stages of integration by partial fractions:
  • During Decomposition: Incorrectly assigning signs when determining the constants (A, B, C, etc.) in the partial fraction expansion, especially when a factor in the denominator is of the form (a - x).
  • During Integration: Forgetting the negative sign that arises when integrating terms like 1/(a - bx). The integral of 1/(ax + b) dx is (1/a)ln|ax+b| + C, but for 1/(a - bx) dx, it is (-1/b)ln|a-bx| + C. This is a very common oversight.
💭 Why This Happens:
These errors primarily stem from a lack of careful attention to detail and haste during calculations. When equating coefficients or substituting values to find constants, a minor algebraic slip can propagate. During integration, students often apply the standard ln|denominator| rule without considering the coefficient of 'x', particularly when it's negative, leading to missing negative signs.
✅ Correct Approach:
Always be meticulous with algebraic signs during the decomposition step. Double-check your equations when finding the constants. More importantly, when integrating terms of the form 1/(ax + b) or 1/(a - bx), remember the general formula for ∫ 1/(Ax + B) dx = (1/A)ln|Ax + B| + C. If A is negative, ensure the negative sign is correctly introduced.
📝 Examples:
❌ Wrong:
Consider the integral ∫ 1/(x(1-x)) dx.
Incorrect thought process:
Partial fraction decomposition: 1/(x(1-x)) = A/x + B/(1-x).
Solving for A and B gives A=1, B=1.
So, ∫ (1/x + 1/(1-x)) dx.
Student might incorrectly write: ln|x| + ln|1-x| + C.
Error: The integral of 1/(1-x) is not ln|1-x|.
✅ Correct:
For the integral ∫ 1/(x(1-x)) dx:
Correct Partial Fraction Decomposition:
1/(x(1-x)) = 1/x + 1/(1-x)
Correct Integration:
∫ (1/x + 1/(1-x)) dx
= ∫ (1/x) dx + ∫ (1/(1-x)) dx
= ln|x| + (-1/1)ln|1-x| + C
= ln|x| - ln|1-x| + C
= ln|x/(1-x)| + C
Key takeaway: The coefficient of x in (1-x) is -1, which introduces the negative sign.
💡 Prevention Tips:
  • JEE Tip: Always perform a quick mental check or a derivative of your integrated result to ensure it matches the integrand.
  • Coefficient Check: Before integrating 1/(Ax + B), explicitly identify the coefficient A (which can be positive or negative) and include 1/A in your result.
  • Careful Substitution: When substituting values for 'x' to find constants A, B, etc., pay extra attention to how signs propagate in the equations.
  • Practice: Regular practice with problems involving factors like (a - bx) will build familiarity and reduce errors.
JEE_Main
Minor Approximation

❌ <span style='color: #FF0000;'>Incorrect Approximation of Partial Fraction Form for Repeated Factors</span>

Students often 'approximate' the partial fraction decomposition by not including all necessary terms when dealing with repeated linear factors or irreducible quadratic factors in the denominator. They might miss the higher powers of repeated factors or use an incorrect numerator form for irreducible quadratics (e.g., A/(ax^2+bx+c) instead of (Ax+B)/(ax^2+bx+c)). This represents an approximation of the correct algebraic structure, leading to an incomplete and ultimately incorrect decomposition.
💭 Why This Happens:
  • Lack of a thorough understanding of the rules for different types of factors in the denominator.
  • Haste or carelessness during the setup phase, assuming a simpler form is sufficient.
  • Focusing only on distinct factors and overlooking their multiplicity (how many times they are repeated).
✅ Correct Approach:
Always meticulously identify the type of factors in the denominator (linear, repeated linear, irreducible quadratic, repeated irreducible quadratic) and apply the correct corresponding partial fraction form. For a repeated linear factor (ax+b)^n, you must include terms for all powers up to n: A1/(ax+b) + A2/(ax+b)2 + ... + An/(ax+b)n. For an irreducible quadratic factor (ax2+bx+c), the term must be (Ax+B)/(ax2+bx+c).
📝 Examples:
❌ Wrong:
For the integral ∫ (x+2) / (x(x-1)2) dx, a common incorrect 'approximation' for the partial fraction decomposition is:
(x+2) / (x(x-1)2) ≈ A/x + B/(x-1)
This form incorrectly approximates the necessary terms by omitting the C/(x-1)2 term required for the repeated factor (x-1)2. This 'approximation' will lead to incorrect coefficients and an ultimately wrong integral.
✅ Correct:
For the integral ∫ (x+2) / (x(x-1)2) dx, the correct partial fraction decomposition is:
(x+2) / (x(x-1)2) = A/x + B/(x-1) + C/(x-1)2
After finding the constants (A=2, B=-2, C=3), the integral becomes:
∫ [2/x - 2/(x-1) + 3/(x-1)2] dx = 2ln|x| - 2ln|x-1| - 3/(x-1) + C
💡 Prevention Tips:
  • Systematic Factor Analysis: Before starting decomposition, carefully list all factors in the denominator, noting their type (linear, quadratic) and their multiplicity (whether they are repeated and how many times).
  • Memorize Forms: Ensure you clearly remember the standard partial fraction forms for each type of factor (linear, repeated linear, irreducible quadratic). A small reference table can be very helpful.
  • Double-Check Structure: After writing down the assumed partial fraction form, compare it against the original denominator to ensure every power of every factor is accurately accounted for.
  • JEE Focus: In JEE (and CBSE), partial fractions are often a crucial intermediate step in a larger integration problem. An error in decomposition will propagate, making accuracy paramount. There is no room for approximation in the algebraic decomposition itself.
JEE_Main
Minor Other

❌ Ignoring Improper Rational Functions Before Partial Fractions

A common oversight is to directly apply partial fraction decomposition to rational functions where the degree of the numerator is greater than or equal to the degree of the denominator. This is incorrect, as partial fraction decomposition is only valid for proper rational functions (where the degree of the numerator is strictly less than the degree of the denominator).
💭 Why This Happens:
Students often rush to identify the type of factors in the denominator (linear, quadratic, repeated) without first checking the fundamental condition for applying partial fractions. This usually stems from a lack of clarity on the prerequisites for the method or simply overlooking this crucial initial step under exam pressure.
✅ Correct Approach:
Before attempting partial fraction decomposition, always compare the degrees of the numerator and denominator. If the degree of the numerator is greater than or equal to the degree of the denominator (i.e., it's an improper rational function), you must first perform polynomial long division. This process will transform the improper fraction into a sum of a polynomial and a proper rational function. The polynomial part can be integrated directly, and the proper rational function can then be decomposed using partial fractions.
📝 Examples:
❌ Wrong:
Directly attempting to decompose $frac{x^3 + 2x + 1}{x^2 - 1}$ into partial fractions without long division. For instance, assuming $frac{x^3 + 2x + 1}{x^2 - 1} = frac{A}{x-1} + frac{B}{x+1}$ is fundamentally incorrect.
✅ Correct:
To integrate $int frac{x^3 + 2x + 1}{x^2 - 1} dx$:
1. Perform polynomial long division:
$(x^3 + 2x + 1) div (x^2 - 1) = x ext{ with a remainder of } (3x + 1)$.
So, $frac{x^3 + 2x + 1}{x^2 - 1} = x + frac{3x + 1}{x^2 - 1}$.
2. Now, decompose the proper rational function: $frac{3x + 1}{x^2 - 1} = frac{3x + 1}{(x-1)(x+1)} = frac{A}{x-1} + frac{B}{x+1}$.
Solving for A and B gives $A=2, B=1$.
3. The integral becomes $int left(x + frac{2}{x-1} + frac{1}{x+1}
ight) dx$.
Integrating term by term yields $frac{x^2}{2} + 2 ln|x-1| + ln|x+1| + C$.
💡 Prevention Tips:
  • Check Degree First: Always compare $ ext{deg(Numerator)}$ and $ ext{deg(Denominator)}$ as the very first step in any partial fraction problem.
  • Polynomial Long Division: If $ ext{deg(Numerator)} ge ext{deg(Denominator)}$, perform long division. This applies to both CBSE and JEE problems.
  • Practice: Solve a variety of problems, including those where long division is required, to build this habit.
  • JEE Tip: In competitive exams, questions might intentionally include improper fractions to test this foundational understanding. Don't fall for the trap of jumping straight to decomposition!
JEE_Main
Minor Approximation

❌ Ignoring Improper Fractions Before Decomposition

A common minor mistake is attempting to apply partial fraction decomposition directly to a rational function that is improper. A rational function P(x)/Q(x) is improper if the degree of the numerator P(x) is greater than or equal to the degree of the denominator Q(x). Partial fraction decomposition is only directly applicable to proper rational functions, where the degree of the numerator is strictly less than the degree of the denominator.
💭 Why This Happens:
This error often stems from an incomplete understanding of the prerequisites for partial fraction decomposition. Students might rush into applying the standard forms without first checking the degrees of the polynomials. It's an oversight of a foundational step, sometimes due to a superficial 'approximation' of the problem's structure rather than a rigorous check.
✅ Correct Approach:
Before applying partial fraction decomposition, always check the degrees of the numerator and the denominator. If the fraction is improper, perform polynomial long division to express it as a sum of a polynomial and a proper rational function. Then, apply partial fraction decomposition only to the resulting proper rational part. This step is crucial for both CBSE and JEE exams.
📝 Examples:
❌ Wrong:
Consider $int frac{x^2 + 3x + 2}{x^2 - 4} dx$.
Incorrect start: Directly assume $frac{x^2 + 3x + 2}{x^2 - 4} = frac{A}{x-2} + frac{B}{x+2}$. This is wrong because deg(Numerator) = deg(Denominator) = 2, making it an improper fraction. Setting up this way will lead to incorrect values for A and B or an unsolvable system.
✅ Correct:
For $int frac{x^2 + 3x + 2}{x^2 - 4} dx$:
Step 1: Check degrees. Deg(Numerator) = 2, Deg(Denominator) = 2. It's an improper fraction.
Step 2: Perform polynomial long division.
$frac{x^2 + 3x + 2}{x^2 - 4} = frac{(x^2 - 4) + 3x + 6}{x^2 - 4} = 1 + frac{3x + 6}{x^2 - 4}$
Step 3: Apply partial fractions to the proper part. Now, decompose $frac{3x + 6}{x^2 - 4} = frac{3x + 6}{(x-2)(x+2)}$ into $frac{A}{x-2} + frac{B}{x+2}$.
Then, the original integral becomes $int left(1 + frac{A}{x-2} + frac{B}{x+2}
ight) dx$.
💡 Prevention Tips:
  • Always Verify Degrees: Make it a habit to compare the degree of the numerator with the degree of the denominator before any other step.
  • Master Polynomial Long Division: Ensure you are proficient in polynomial long division, as it's a fundamental prerequisite.
  • Understand the 'Why': Remember that partial fraction decomposition aims to simplify proper rational functions into simpler fractions that are easier to integrate. Improper fractions require an initial simplification.
  • Practice Regularly: Work through problems that involve both proper and improper fractions to solidify this concept.
CBSE_12th
Minor Sign Error

❌ Sign Errors in Coefficient Determination (Minor Severity)

Students frequently make minor sign errors when determining the values of the coefficients (A, B, C, etc.) during the partial fraction decomposition. This often happens either when substituting specific values of 'x' to find coefficients or when comparing coefficients of like powers of 'x' from both sides of the equation. A single misplaced negative sign can lead to incorrect partial fractions and, consequently, an incorrect final integral.
💭 Why This Happens:
This minor error typically stems from:
  • Carelessness during substitution: Incorrectly handling negative numbers when plugging in values for 'x' (e.g., `-1-3 = -4` might become `4`).
  • Distributive property oversight: Forgetting to distribute a negative sign across all terms inside a parenthesis (e.g., `-(x-1)` becoming `-x-1` instead of `-x+1`).
  • Rushed calculations: Performing mental arithmetic too quickly without proper verification.
  • Basic algebraic slips: Simple addition/subtraction errors involving negative numbers.
✅ Correct Approach:
To avoid sign errors, follow these steps meticulously:
  • Isolate and substitute carefully: When substituting values of 'x' that make a term zero, write down each step clearly.
  • Check distributive property: Always double-check expansion of terms involving negative signs.
  • Systematic comparison: If comparing coefficients, align terms with the same power of 'x' before equating their coefficients. Pay close attention to the signs of each term.
  • Verification: After finding A, B, C, quickly substitute them back into the expanded original equation for a simple cross-check, or test with another 'x' value.
📝 Examples:
❌ Wrong:
Consider the decomposition of (x-3) / (x(x+1)).
Let (x-3) / (x(x+1)) = A/x + B/(x+1)
x-3 = A(x+1) + Bx
To find A, let x=0: 0-3 = A(0+1) + B(0)
Wrong: Student mistakenly writes 3 = A (dropping the negative sign).
So, A = 3 and then proceeds to find B, leading to an incorrect decomposition.
✅ Correct:
For the same decomposition of (x-3) / (x(x+1)):
x-3 = A(x+1) + Bx
To find A, let x=0:
0-3 = A(0+1) + B(0)
-3 = A(1)
A = -3 (Correct calculation of A)
To find B, let x=-1:
-1-3 = A(-1+1) + B(-1)
-4 = 0 - B
B = 4 (Correct calculation of B)
The correct partial fraction decomposition is -3/x + 4/(x+1).
💡 Prevention Tips:
  • Write down intermediate steps: Especially when dealing with negative numbers, avoid doing too many steps mentally.
  • Use parentheses diligently: Always use parentheses when substituting negative values or when distributing a negative sign.
  • Double-check signs: After each line of calculation, quickly review the signs of all terms.
  • Practice consistently: Regular practice with diverse problems helps in building a strong intuition for handling signs correctly. For CBSE exams, neatness and step-by-step clarity are crucial for avoiding such errors and for fetching marks even if the final answer has a minor slip.
CBSE_12th
Minor Unit Conversion

❌ Ignoring Improper Rational Functions Before Partial Fraction Decomposition

Students often fail to recognize when a rational function is improper (degree of numerator ≥ degree of denominator) and attempt to apply partial fraction decomposition directly. This is a fundamental error because partial fractions are only applicable to proper rational functions.
💭 Why This Happens:
This mistake stems from a lack of thorough understanding of the prerequisites for applying partial fraction method. Students may rush into setting up the decomposition without first verifying the degrees of the numerator and denominator, or they might simply forget the necessary step of polynomial long division.
✅ Correct Approach:
Always first compare the degree of the numerator (N) with the degree of the denominator (D). If deg(N) ≥ deg(D), perform polynomial long division to express the improper rational function as a sum of a polynomial and a proper rational function. Then, apply partial fraction decomposition only to the resulting proper rational function.
📝 Examples:
❌ Wrong:
For the integral ∫ (x³ + x + 1) / (x² - 1) dx, a common mistake is to directly assume a decomposition like A/(x-1) + B/(x+1) for the entire function, which is incorrect.
✅ Correct:
To correctly solve ∫ (x³ + x + 1) / (x² - 1) dx:
1. Recognize it's an improper fraction (deg(N)=3, deg(D)=2).
2. Perform polynomial long division:
(x³ + x + 1) / (x² - 1) = x + (2x + 1) / (x² - 1)
3. Now, apply partial fractions only to the proper rational part: (2x + 1) / (x² - 1) = (2x + 1) / ((x-1)(x+1)).
4. Decompose: (2x + 1) / ((x-1)(x+1)) = A/(x-1) + B/(x+1).
5. The original integral becomes ∫ [x + A/(x-1) + B/(x+1)] dx.
💡 Prevention Tips:
  • Always check degrees: Make it a habit to compare the degrees of the numerator and denominator as the very first step.
  • Polynomial Long Division: Practice polynomial long division to perform this 'conversion' efficiently.
  • Prerequisite Knowledge: Revisit the conditions for applying partial fraction decomposition (only for proper rational functions).
CBSE_12th
Minor Formula

❌ Incorrect Partial Fraction Form for Repeated Linear Factors

Students frequently make an error in setting up the partial fraction decomposition when the denominator contains repeated linear factors. Instead of including terms for each power of the repeated factor, they often only account for the highest power or simply treat it as a distinct single factor.
💭 Why This Happens:
This mistake typically arises from an incomplete understanding of the theoretical basis of partial fractions. Students might simplify the decomposition forms, forgetting that each power of a repeated factor (e.g., (x-a), (x-a)² , ..., (x-a)ⁿ) needs a separate term with an independent constant numerator to fully represent all possible combinations when fractions are re-combined.
✅ Correct Approach:
For a repeated linear factor of the form (ax + b)ⁿ in the denominator, the partial fraction decomposition must include 'n' distinct terms, each with an increasing power of the factor in the denominator and a constant in the numerator. The general form is:
A₁/(ax + b) + A₂/(ax + b)² + ... + Aₙ/(ax + b)ⁿ
📝 Examples:
❌ Wrong:
For the integral ∫ x / ((x-1)²(x+2)) dx,
An incorrect setup would be:
A/(x-1)² + B/(x+2)
This misses the A/(x-1) term.
✅ Correct:
For the integral ∫ x / ((x-1)²(x+2)) dx,
The correct partial fraction decomposition is:
A/(x-1) + B/(x-1)² + C/(x+2)
Each power of (x-1) up to the highest power is represented.
💡 Prevention Tips:
  • Master the Forms: Dedicate time to thoroughly understand and memorize all standard partial fraction decomposition forms, especially for repeated linear factors and irreducible quadratic factors.
  • Practice Decomposition: Before proceeding to integration, explicitly write down the partial fraction decomposition for various expressions. This builds confidence and reinforces correct formula application.
  • Cross-Check Terms: Always verify that the number of terms in your decomposition matches the complexity of the denominator. For a factor (ax+b)ⁿ, ensure there are 'n' corresponding terms.
CBSE_12th
Minor Calculation

❌ Arithmetic Errors in Solving for Partial Fraction Coefficients

Students often make minor arithmetic mistakes while solving for the unknown coefficients (A, B, C, etc.) in partial fraction decomposition. These errors can occur during the substitution method (setting x to specific roots) or the coefficient comparison method (equating coefficients of like powers of x). Even a small sign error or calculation slip can lead to incorrect coefficient values, rendering the final integral incorrect.
💭 Why This Happens:
This minor error usually stems from:
  • Rushing: Students often hurry through the algebraic steps to save time.
  • Carelessness: Lack of attention to detail, especially with negative signs or fractions.
  • Lack of double-checking: Not verifying the calculated coefficients by substituting them back or using an alternative method.
✅ Correct Approach:
Always approach the calculation of coefficients systematically and with careful attention to arithmetic.
  • For Substitution Method: Substitute values of x corresponding to the roots of the denominator factors. Perform each multiplication and addition step meticulously.
  • For Coefficient Comparison Method: Expand the expression fully, group terms by powers of x, and then carefully equate coefficients. Be extra vigilant with negative signs.
  • Verification: If time permits, cross-check one or two coefficients using both methods or by substituting one random value of x (not a root) into the decomposed form and the original expression to see if they match.
📝 Examples:
❌ Wrong:
Consider the integral of (frac{x+1}{(x-1)(x+2)}).
Decomposition: (frac{x+1}{(x-1)(x+2)} = frac{A}{x-1} + frac{B}{x+2})
(x+1 = A(x+2) + B(x-1))
Wrong Calculation for A: Set (x=1)
(1+1 = A(1+2) + B(1-1))
(2 = A(3))
(A = frac{3}{2}) (Incorrect)
✅ Correct:
For the same integral of (frac{x+1}{(x-1)(x+2)}).
(x+1 = A(x+2) + B(x-1))
Correct Calculation for A: Set (x=1)
(1+1 = A(1+2) + B(1-1))
(2 = A(3) + B(0))
(2 = 3A)
(A = frac{2}{3}) (Correct)
Correct Calculation for B: Set (x=-2)
(-2+1 = A(-2+2) + B(-2-1))
(-1 = A(0) + B(-3))
(-1 = -3B)
(B = frac{1}{3}) (Correct)
💡 Prevention Tips:
  • Write down all steps: Avoid mental calculations for crucial steps.
  • Double-check signs: Be extremely careful with positive and negative signs, especially when multiplying or substituting negative numbers.
  • Practice consistently: Regular practice of partial fraction problems will improve speed and accuracy in coefficient determination.
  • Verify with a quick check: After finding A and B, pick a simple value for x (e.g., x=0, if not a root) and ensure that (frac{x+1}{(x-1)(x+2)}) equals (frac{A}{x-1} + frac{B}{x+2}) for that x.
CBSE_12th
Minor Conceptual

❌ Ignoring Improper Fractions Before Partial Decomposition

Students frequently attempt to apply partial fraction decomposition directly to improper rational functions where the degree of the numerator is greater than or equal to the degree of the denominator. The method of partial fractions is strictly applicable only to proper rational functions.
💭 Why This Happens:
This mistake stems from a fundamental misunderstanding of the prerequisite condition for partial fraction decomposition. Students often rush to identify the factors of the denominator without first checking the degrees of the polynomials, leading to incorrect setup and ultimately wrong integration.
✅ Correct Approach:
Before applying partial fraction decomposition, always ensure the given rational function is a proper fraction (degree of numerator < degree of denominator). If it's an improper fraction, perform polynomial long division first. This converts the improper fraction into a sum of a polynomial and a proper rational function. Then, apply partial fraction decomposition only to the resulting proper fraction part.
📝 Examples:
❌ Wrong:

Consider the integral: (int frac{x^2 + 3x + 2}{x^2 - 1} dx)

Wrong Approach: Directly assuming (frac{x^2 + 3x + 2}{(x-1)(x+1)} = frac{A}{x-1} + frac{B}{x+1}). This is incorrect because the degree of the numerator (2) is equal to the degree of the denominator (2).

✅ Correct:

Consider the integral: (int frac{x^2 + 3x + 2}{x^2 - 1} dx)

Correct Approach:

  1. Perform Polynomial Long Division:
    (frac{x^2 + 3x + 2}{x^2 - 1} = frac{(x^2 - 1) + 3x + 3}{x^2 - 1} = 1 + frac{3x + 3}{x^2 - 1})
  2. Decompose the Proper Fraction: Now, apply partial fractions to (frac{3x + 3}{x^2 - 1}).
    (frac{3x + 3}{(x-1)(x+1)} = frac{A}{x-1} + frac{B}{x+1})
    Multiply by ((x-1)(x+1)): (3x + 3 = A(x+1) + B(x-1))
    Set (x=1): (3(1)+3 = A(1+1) implies 6 = 2A implies A=3)
    Set (x=-1): (3(-1)+3 = B(-1-1) implies 0 = -2B implies B=0)
    So, (frac{3x + 3}{x^2 - 1} = frac{3}{x-1})
  3. Integrate:
    (int left(1 + frac{3}{x-1}
    ight) dx = x + 3 ln|x-1| + C)
💡 Prevention Tips:
  • Always Check Degrees First: Before attempting any partial fraction decomposition, compare the degree of the numerator with that of the denominator.
  • CBSE & JEE Reminder: This is a fundamental step. Skipping long division for improper fractions is a common error that leads to zero marks for the decomposition part.
  • Practice Long Division: Ensure you are proficient with polynomial long division, as it's a crucial prerequisite for such problems.
  • Revisit Definitions: Understand the difference between proper and improper rational functions thoroughly.
CBSE_12th
Minor Conceptual

❌ Incorrect Partial Fraction Decomposition for Repeated Linear Factors

A common conceptual error is incorrectly setting up the partial fraction decomposition when the denominator contains repeated linear factors. Students often treat a repeated factor, such as (ax+b)n, as a simple linear factor, leading to an insufficient number of terms in the decomposition.
💭 Why This Happens:
This mistake stems from an incomplete understanding of the rules for partial fraction decomposition. Students might correctly identify linear factors but overlook their multiplicity (the power to which they are raised), failing to generate all necessary terms for the repeated factor. Rushing through the setup is a frequent contributor.
✅ Correct Approach:
When a denominator contains a repeated linear factor (ax+b)n, it requires 'n' individual partial fraction terms. These terms should have denominators corresponding to each power of the factor, from 1 up to 'n'.
For (ax+b)n, the correct form is:
$$frac{A_1}{(ax+b)} + frac{A_2}{(ax+b)^2} + dots + frac{A_n}{(ax+b)^n}$$
📝 Examples:
❌ Wrong:
Consider the integral of $$frac{3x+1}{(x-2)^2(x+1)}$$.
An incorrect partial fraction setup would be:
$$frac{A}{(x-2)} + frac{B}{(x+1)}$$
This misses the term corresponding to the second power of (x-2).
✅ Correct:
For the integral of $$frac{3x+1}{(x-2)^2(x+1)}$$, the correct partial fraction decomposition is:
$$frac{A}{(x-2)} + frac{B}{(x-2)^2} + frac{C}{(x+1)}$$
Here, two distinct terms are used for the repeated factor (x-2)2, along with one term for the simple linear factor (x+1).
💡 Prevention Tips:
  • Factorization Check: Always completely factorize the denominator first, paying close attention to any factors raised to a power greater than one.
  • JEE Tip: Multiplicity Rule: For every factor (ax+b)n in the denominator, ensure you write 'n' terms in your partial fraction setup, with denominators (ax+b), (ax+b)2, ..., (ax+b)n.
  • Practice: Work through problems involving various types of repeated factors (linear and irreducible quadratic) to solidify the decomposition rules.
JEE_Advanced
Minor Calculation

❌ <span style='color:#FF0000;'>Algebraic Errors in Determining Partial Fraction Coefficients</span>

Students frequently make minor algebraic slips while solving for the unknown constants (A, B, C, etc.) in the partial fraction decomposition. These errors, often involving incorrect signs or arithmetic, lead to incorrect values for the coefficients, thereby making the final integrated expression erroneous.
💭 Why This Happens:
  • Haste and lack of carefulness in solving the system of equations for the coefficients.
  • Sign errors during substitution of 'x' values or when transposing terms across the equality.
  • Miscalculation during basic arithmetic operations like multiplication or subtraction.
  • Failure to cross-check the derived values of the constants.
✅ Correct Approach:
  • Systematic Substitution: Substitute the roots of the denominator into the identity to directly find as many coefficients as possible.
  • Equating Coefficients: For remaining coefficients, systematically equate the coefficients of like powers of 'x' on both sides of the identity.
  • Double-Check (JEE Tip): Always verify your calculated coefficients. A quick way is to substitute a convenient 'x' value (not one of the roots used previously) into the original partial fraction identity to see if both sides match.
📝 Examples:
❌ Wrong:
Consider the integral `(I = int frac{3x+2}{(x-1)(x+2)} dx)`.
The partial fraction decomposition form is `(frac{3x+2}{(x-1)(x+2)} = frac{A}{x-1} + frac{B}{x+2})`.
Multiplying by `((x-1)(x+2))` gives the identity: `(3x+2 = A(x+2) + B(x-1))`.
Incorrect Calculation:
  1. Substitute `(x=1)`: `(3(1)+2 = A(1+2) implies 5 = 3A implies A = 5/3)` (Correct).
  2. Substitute `(x=-2)`: `(3(-2)+2 = B(-2-1) implies -6+2 = B(-3) implies -4 = -3B)`
    Common Mistake: A student might erroneously write `(-4 = 3B)` due to a sign error, leading to `(B = -4/3)`.
Using `(A=5/3)` and incorrect `(B=-4/3)` would lead to an incorrect integrated result: `(frac{5}{3}ln|x-1| - frac{4}{3}ln|x+2| + C)`.
✅ Correct:
Using the same integral `(I = int frac{3x+2}{(x-1)(x+2)} dx)` and the identity `(3x+2 = A(x+2) + B(x-1))`:
  1. Substitute `(x=1)`: `(3(1)+2 = A(1+2) implies 5 = 3A implies A = 5/3)`.
  2. Substitute `(x=-2)`: `(3(-2)+2 = B(-2-1) implies -4 = -3B implies B = 4/3)`.
The correct values are `(A = 5/3)` and `(B = 4/3)`. The correct integral is: `(int (frac{5/3}{x-1} + frac{4/3}{x+2}) dx = frac{5}{3}ln|x-1| + frac{4}{3}ln|x+2| + C)`.
Cross-check (e.g., for x=0): `(3(0)+2 = A(0+2) + B(0-1) implies 2 = 2A - B)`. Substituting `(A=5/3)` and `(B=4/3)`: `(2(5/3) - 4/3 = 10/3 - 4/3 = 6/3 = 2)`. The values match, confirming correctness.
💡 Prevention Tips:
  • Work deliberately: Take your time with algebraic steps; avoid rushing through calculations.
  • Use ample space: Prevent cramped calculations by utilizing enough rough work space.
  • Re-check signs: Pay close attention to signs at every step, especially when multiplying by negative numbers or transposing terms.
  • Cross-verify (JEE Advanced Specific): After finding all coefficients, substitute them back into the original partial fraction identity and test with a simple 'x' value (different from the roots used) to confirm consistency. This can save valuable marks!
JEE_Advanced
Minor Formula

❌ Incorrect Partial Fraction Decomposition Form

Students frequently make errors in setting up the correct form of partial fraction decomposition, particularly for denominators containing repeated linear factors or irreducible quadratic factors. This foundational error leads to an entirely incorrect integration process.
💭 Why This Happens:
This mistake stems from a superficial understanding of the rules for different types of factors. Many students rote-memorize without grasping the underlying logic, leading to confusion when faced with complex denominators common in JEE Advanced. Sometimes, they also forget to perform polynomial long division if the numerator's degree is greater than or equal to the denominator's degree.
✅ Correct Approach:
Always identify the nature of each factor in the denominator and apply the corresponding partial fraction form rigorously. Before decomposition, ensure that the degree of the numerator is strictly less than the degree of the denominator. If not, perform polynomial long division first.
📝 Examples:
❌ Wrong:
Consider the integral of ∫ (x+1) / (x(x-1)^2 (x^2+4)) dx.
A common wrong decomposition for the last two factors would be:
A/x + B/(x-1) + C/(x-1)^2 + D/(x^2+4)
Here, D/(x^2+4) is incorrect; for an irreducible quadratic factor like x^2+4, the numerator must be linear (Dx+E).
✅ Correct:
For the same integral ∫ (x+1) / (x(x-1)^2 (x^2+4)) dx,
The correct partial fraction decomposition should be:
A/x + B/(x-1) + C/(x-1)^2 + (Dx+E)/(x^2+4)
Each term correctly accounts for the type of factor in the denominator, ensuring a solvable system of equations for the constants.
💡 Prevention Tips:
  • Master the Standard Forms: Thoroughly understand and practice setting up the five standard types of partial fraction decompositions (distinct linear, repeated linear, irreducible quadratic).
  • Degree Check: Always verify that the degree of the numerator is less than the degree of the denominator before starting decomposition.
  • Practice Complex Problems: Focus on problems involving both repeated linear and irreducible quadratic factors, as these are common pitfalls in JEE Advanced.
  • JEE vs. CBSE: While CBSE generally sticks to simpler forms, JEE Advanced frequently combines multiple complex factors, demanding precise application of decomposition rules.
JEE_Advanced
Minor Unit Conversion

❌ Incorrect Algebraic Setup for Partial Fraction Decomposition

Students frequently make errors in correctly setting up the partial fraction form of a rational function before integrating. These errors are purely algebraic and structural. Important Note: The concept of 'Unit Conversion understanding' is not applicable to Integration by Partial Fractions, as this topic deals with mathematical functions and algebraic manipulation, not physical quantities or units.
💭 Why This Happens:
This mistake primarily stems from an incomplete understanding or misapplication of the rules for different types of denominator factors (linear, repeated linear, irreducible quadratic). Rushing the initial setup or failing to check if the rational function is proper also contributes to these errors. In JEE Advanced, such algebraic slips can lead to completely wrong integrals.
✅ Correct Approach:
Always ensure the rational function is a proper fraction (degree of numerator < degree of denominator) before decomposition. If not, perform polynomial long division first. Then, systematically apply the correct decomposition rules based on the nature of the denominator's factors:

  • For each non-repeated linear factor (ax+b), use A/(ax+b).

  • For each repeated linear factor (ax+b)n, use A1/(ax+b) + A2/(ax+b)2 + ... + An/(ax+b)n.

  • For each non-repeated irreducible quadratic factor (ax2+bx+c), use (Ax+B)/(ax2+bx+c).

  • For each repeated irreducible quadratic factor (ax2+bx+c)n, use (A1x+B1)/(ax2+bx+c) + ... + (Anx+Bn)/(ax2+bx+c)n.

📝 Examples:
❌ Wrong:
For integrating $int frac{x^2}{(x-1)(x^2+1)} dx$, a common incorrect setup for partial fractions is:
$,qquad frac{A}{x-1} + frac{B}{x^2+1}$
This is incorrect because $x^2+1$ is an irreducible quadratic factor, which requires a linear term in its numerator.
✅ Correct:
The correct partial fraction decomposition for $frac{x^2}{(x-1)(x^2+1)}$ should be:
$,qquad frac{A}{x-1} + frac{Bx+C}{x^2+1}$
Once this form is correctly established, solving for A, B, C, and then integrating becomes a standard process.
💡 Prevention Tips:

  • Master Decomposition Rules: Dedicate time to thoroughly understand and memorize the specific rules for setting up partial fractions for all types of denominator factors.

  • Check for Proper Fraction: Always perform polynomial long division if the degree of the numerator is greater than or equal to the degree of the denominator before attempting decomposition.

  • Practice Variety: Work through numerous examples from JEE Advanced past papers covering linear, repeated linear, and irreducible quadratic factors to build proficiency.

  • Double-Check Setup: Before proceeding to solve for the constants, quickly review your partial fraction setup against the original rational function's denominator to catch errors early.

JEE_Advanced
Minor Sign Error

❌ Sign Errors in Determining Coefficients of Partial Fractions

Students frequently make sign errors when calculating the coefficients (A, B, C, etc.) during the partial fraction decomposition step. This often happens due to careless algebraic manipulation, especially when substituting values to solve for constants or when combining terms. A single sign error here propagates through the entire integration process, leading to an incorrect final answer.
💭 Why This Happens:
  • Rushed Calculations: Under exam pressure, students often rush algebraic steps, overlooking negative signs.
  • Complex Substitutions: When substituting negative values for 'x' or dealing with expressions involving subtractions, sign mistakes are common.
  • Lack of Double-Checking: Not verifying the calculated coefficients by cross-multiplication or by simply re-evaluating the steps.
  • Distribution Errors: Incorrectly distributing a negative sign across a parenthesis.
✅ Correct Approach:
Always be meticulous with algebraic steps. Use parentheses liberally, especially when substituting negative numbers or distributing negative signs. After finding the coefficients, it's good practice to quickly combine the decomposed fractions back to the original form to verify the coefficients and their signs before proceeding with integration. For JEE Advanced, precision in calculation is paramount.
📝 Examples:
❌ Wrong:
Consider decomposing (x-5) / (x² - 4x + 3) for integration.
Let (x-5) / ((x-1)(x-3)) = A/(x-1) + B/(x-3)
This leads to: x-5 = A(x-3) + B(x-1)
To find A, set x=1: 1-5 = A(1-3) => -4 = -2A => A = 2
To find B, set x=3: 3-5 = B(3-1)
Wrong step: A student might carelessly write 2 = 2B instead of -2 = 2B, leading to B = 1.
Thus, the incorrect decomposition becomes: 2/(x-1) + 1/(x-3)
Integrating this yields: 2ln|x-1| + ln|x-3| + C (Incorrect)
✅ Correct:
Using the same problem: (x-5) / (x² - 4x + 3)
We have: x-5 = A(x-3) + B(x-1)
For A: Set x=1: 1-5 = A(1-3) => -4 = -2A => A = 2
For B: Set x=3: 3-5 = B(3-1) => -2 = 2B => B = -1
The correct decomposition is: 2/(x-1) + (-1)/(x-3) or 2/(x-1) - 1/(x-3)
Integrating this yields: ∫ (2/(x-1) - 1/(x-3)) dx = 2ln|x-1| - ln|x-3| + C (Correct)
💡 Prevention Tips:
  • Use Parentheses: Always use parentheses when substituting values, especially negative ones, to avoid sign confusion.
  • Verify Algebra: After finding coefficients, quickly cross-check the algebraic steps. Even a mental re-evaluation can catch errors.
  • Systematic Approach: Follow a consistent method for solving coefficients (e.g., substitution method or comparing coefficients) to minimize errors.
  • Intermediate Checks: For CBSE students, show all steps clearly. For JEE Advanced, while speed is crucial, accuracy of signs should never be compromised.
JEE_Advanced
Minor Approximation

❌ Incorrect Numerator Form for Irreducible Quadratic Factors

A common mistake, particularly in JEE Advanced, is to incorrectly set up the numerator for an irreducible quadratic factor in the denominator during partial fraction decomposition. Students often assume it should be a constant (like 'A' or 'B'), similar to linear factors, rather than a linear expression (Ax + B). This fundamental error in decomposition setup leads to an entirely incorrect integration.
💭 Why This Happens:
This misunderstanding often arises from overgeneralizing the rule for linear factors, where the numerator is indeed a constant. Students might inadvertently 'approximate' the required linear numerator (Ax + B) to a simpler constant, especially when under exam pressure or if their foundational understanding of partial fraction forms for different factor types is weak. They simplify the form without considering the underlying algebraic necessity.
✅ Correct Approach:
For every irreducible quadratic factor of the form (ax² + bx + c) in the denominator (where the discriminant b² - 4ac < 0, meaning it cannot be factored into real linear factors), the corresponding numerator in the partial fraction decomposition must be a linear expression, i.e., (Ax + B). This ensures that the algebraic structure of the original rational function is correctly represented.
📝 Examples:
❌ Wrong:
For the integral ∫ (3x² + 2) / [(x - 2)(x² + x + 1)] dx, a student might incorrectly set up the partial fraction decomposition as:
A/(x - 2) + B/(x² + x + 1)
Here, `x² + x + 1` is an irreducible quadratic factor (since 1² - 4(1)(1) = -3 < 0). The numerator 'B' for this factor is incorrect.
✅ Correct:
The correct partial fraction decomposition for ∫ (3x² + 2) / [(x - 2)(x² + x + 1)] dx is:
A/(x - 2) + (Bx + C)/(x² + x + 1)
This setup correctly accounts for the irreducible quadratic factor `x² + x + 1` by assigning a linear numerator `(Bx + C)` to it.
💡 Prevention Tips:
  • Categorize Factors First: Before decomposing, always identify if each factor in the denominator is linear, repeated linear, irreducible quadratic, or repeated irreducible quadratic.
  • Memorize Forms: Be absolutely clear on the numerator form for each type of factor. For irreducible quadratic factors, it's always (Ax + B).
  • Check Discriminant: If unsure about a quadratic factor, quickly check its discriminant (b² - 4ac). If it's negative, it's irreducible over real numbers.
  • Practice JEE Advanced Problems: Work through problems that specifically involve irreducible quadratic factors to solidify your understanding and application of the correct form.
JEE_Advanced
Important Formula

❌ Incorrect Decomposition for Repeated Linear Factors

A very common mistake in JEE Advanced is the incorrect setup of the partial fraction decomposition when the denominator contains repeated linear factors. Students often fail to account for all powers of the repeated factor, leading to an incomplete or incorrect decomposition.
💭 Why This Happens:
This error primarily stems from a lack of precise understanding of the partial fraction formula rules. Students might confuse the decomposition for distinct linear factors with that for repeated ones, or simply forget to include terms for all intermediate powers of the repeated factor. Rushing through the setup is another contributing factor.
✅ Correct Approach:
When a linear factor (ax + b) is repeated 'n' times in the denominator (i.e., (ax + b)^n), the partial fraction decomposition must include 'n' separate terms. Each term will have a constant numerator and the denominator will be (ax + b) raised to powers from 1 up to 'n'. For example, for (ax + b)^3, the terms should be A/(ax + b) + B/(ax + b)² + C/(ax + b)³.
📝 Examples:
❌ Wrong:
For the integral ∫ (x+1) / (x(x-1)²) dx, a common incorrect decomposition would be:
(x+1) / (x(x-1)²) = A/x + B/(x-1)
This misses the term for (x-1)², making the entire setup erroneous. The number of unknown constants must match the degree of the denominator polynomial.
✅ Correct:
For the integral ∫ (x+1) / (x(x-1)²) dx, the correct partial fraction decomposition is:
(x+1) / (x(x-1)²) = A/x + B/(x-1) + C/(x-1)²
Here, A, B, and C are constants to be determined. This setup ensures that all powers of the repeated linear factor are accounted for, leading to a solvable system for the constants and a correct integration path.
💡 Prevention Tips:
  • Memorize Rules Thoroughly: Understand and memorize the partial fraction decomposition rules for all cases: distinct linear, repeated linear, distinct irreducible quadratic, and repeated irreducible quadratic factors.
  • Cross-Check Degree: Before proceeding, ensure the number of unknown constants in your decomposition matches the degree of the denominator polynomial.
  • Practice Varied Problems: Solve problems covering all types of factors, paying special attention to repeated factors.
  • JEE Advanced Tip: Errors in decomposition are fundamental and will lead to completely wrong answers. Always double-check your initial partial fraction setup.
JEE_Advanced
Important Unit Conversion

❌ Ignoring Improper Rational Functions – The 'Pre-Conversion' Step

A frequent and critical error in Integration by Partial Fractions is attempting to decompose an improper rational function directly without first converting it into a proper rational function. Students often overlook the prerequisite step of polynomial long division. While the term 'Unit Conversion' isn't applicable to physical units here, this 'conversion of form' is a crucial initial step.
💭 Why This Happens:
This mistake primarily stems from a lack of vigilance in checking the degrees of polynomials or misunderstanding the fundamental condition for applying partial fraction decomposition. Students sometimes forget that partial fractions are strictly applicable only to proper rational functions (where the degree of the numerator is strictly less than the degree of the denominator). Rushing through problems or not recognizing the condition often leads to this error.
✅ Correct Approach:
Before initiating partial fraction decomposition, always compare the degree of the numerator polynomial, P(x), with the degree of the denominator polynomial, Q(x).
  • If deg(P(x)) < deg(Q(x)): The function is a proper rational function; proceed directly with partial fraction decomposition.
  • If deg(P(x)) ≥ deg(Q(x)): The function is an improper rational function. You must perform polynomial long division first. The rational function should be expressed as:
    P(x)/Q(x) = Quotient(x) + Remainder(x)/Q(x)
    Then, apply partial fraction decomposition only to the Remainder(x)/Q(x) term, which is now a proper rational function. Integrate the Quotient(x) (a polynomial) separately.
📝 Examples:
❌ Wrong:
Consider the integral: ∫ (x² + x + 1) / (x + 1) dx
Wrong Approach: Directly try to set (x² + x + 1) / (x + 1) = A / (x + 1). This is incorrect because deg(x² + x + 1) = 2 and deg(x + 1) = 1. The function is improper, and this decomposition form is invalid.
✅ Correct:
For the integral: ∫ (x² + x + 1) / (x + 1) dx
Correct Approach:
  1. Check Degree: deg(Numerator) = 2, deg(Denominator) = 1. Since 2 ≥ 1, it's an improper fraction.
  2. Polynomial Division: Divide (x² + x + 1) by (x + 1).
    (x² + x + 1) = x(x + 1) + 1
    So, (x² + x + 1) / (x + 1) = x + 1 / (x + 1)
  3. Integrate: Now integrate term by term:
    ∫ [x + 1 / (x + 1)] dx = ∫ x dx + ∫ [1 / (x + 1)] dx
    = x²/2 + ln|x + 1| + C
💡 Prevention Tips:
  • Always Check Degrees First: Make it your absolute first step to compare the degrees of the numerator and denominator in any rational function integration problem. This is a fundamental check.
  • Master Polynomial Long Division: Ensure you are proficient and quick in performing polynomial long division accurately, as it's a prerequisite skill.
  • JEE Advanced Alert: Examiners frequently embed improper rational functions to test this foundational understanding. Skipping this 'pre-conversion' step invariably leads to incorrect partial fraction decomposition and a completely wrong final answer.
JEE_Advanced
Important Sign Error

❌ Sign Error in Integrating Terms of Type 1/(a-x)

Students frequently make a sign error when integrating partial fraction terms of the form B/(a-x). After correctly decomposing the rational function into partial fractions, they often integrate ∫ (B/(a-x)) dx as B ln|a-x| + C instead of the correct -B ln|a-x| + C. This is a crucial sign oversight that can lead to completely wrong final answers in JEE Advanced problems.
💭 Why This Happens:
This error primarily stems from:
  • Overlooking the Chain Rule: Students forget that the derivative of (a-x) with respect to x is -1. When applying the integral formula ∫ f'(x)/f(x) dx = ln|f(x)| + C, they neglect the negative factor from the inner derivative.
  • Rushing and Carelessness: In exam pressure, the common form ∫ 1/x dx = ln|x| is generalized without considering the coefficient of x.
  • Lack of Fundamental Clarity: Not thoroughly understanding that ∫ 1/(ax+b) dx = (1/a)ln|ax+b| + C, where 'a' is the coefficient of x. For (a-x), 'a' is -1.
✅ Correct Approach:
Always apply the general formula for integrating linear terms in the denominator:
∫ 1/(ax+b) dx = (1/a)ln|ax+b| + C
When you encounter 1/(a-x), recognize that it can be written as 1/(-1x + a). Here, the coefficient of x is -1. Therefore, its integral is (1/-1)ln|a-x| + C = -ln|a-x| + C.
For a term B/(a-x), the integral is -B ln|a-x| + C. This careful application ensures the correct sign.
📝 Examples:
❌ Wrong:

Consider the integral ∫ (1 / (5-2x)) dx

Incorrect Integration:

∫ (1 / (5-2x)) dx = (1/2)ln|5-2x| + C (Incorrect)
(Mistake: Used 1/2 from the '2x' without considering '-2x')
✅ Correct:

Consider the integral ∫ (1 / (5-2x)) dx

Correct Integration:

∫ (1 / (5-2x)) dx
Here, a = -2 (coefficient of x) and b = 5.
Using ∫ 1/(ax+b) dx = (1/a)ln|ax+b| + C
= (1/-2)ln|5-2x| + C
= -(1/2)ln|5-2x| + C (Correct)
💡 Prevention Tips:
  • Always Identify a: For any term 1/(ax+b), explicitly identify the coefficient a (including its sign).
  • JEE Advanced Focus: Be particularly vigilant with terms like (c-x) or (c-kx), as these are common traps where the negative sign is easily missed.
  • Mental Check: After integrating, quickly differentiate your result mentally (or on scratch paper) to see if you arrive back at the original integrand, paying close attention to signs.
  • Practice Similar Problems: Solve multiple problems involving such terms to build muscle memory for the correct sign.
JEE_Advanced
Important Approximation

❌ <span style='color: #FF0000;'>Incorrectly Applying Partial Fractions to Improper Rational Functions</span>

Students frequently make an 'approximation' that any rational function P(x)/Q(x) can be directly decomposed using partial fractions. This error occurs when the degree of the numerator P(x) is greater than or equal to the degree of the denominator Q(x) (an improper rational function), which violates a fundamental prerequisite for direct application of the method.
💭 Why This Happens:
This mistake stems from a faulty 'approximation understanding' of the scope of partial fraction decomposition. Students often overlook or simply ignore the crucial initial step of checking the degrees of the polynomials, directly jumping to factorizing the denominator and setting up the partial fraction form without performing polynomial long division first.
✅ Correct Approach:
  • Always check the degrees: Before any decomposition, compare deg(P(x)) and deg(Q(x)).
  • Perform Polynomial Long Division: If deg(P(x)) ≥ deg(Q(x)), first divide P(x) by Q(x) to express the rational function as: P(x)/Q(x) = Quotient(x) + Remainder(x)/Q(x).
  • Apply Partial Fractions to Remainder: The Remainder(x)/Q(x) term will now be a proper rational function (deg(Remainder(x)) < deg(Q(x))), to which you can correctly apply partial fraction decomposition.
  • Integrate Separately: The integral becomes ∫ Quotient(x) dx + ∫ [Partial Fractions of Remainder(x)/Q(x)] dx.
📝 Examples:
❌ Wrong:
Integral: ∫ (x3 + x2 + 1)/(x2 - 1) dx
Wrong Start: Directly attempting to decompose (x3 + x2 + 1)/((x-1)(x+1)) into A/(x-1) + B/(x+1). This is incorrect as deg(Numerator) = 3 and deg(Denominator) = 2, making it an improper rational function.
✅ Correct:
Integral: ∫ (x3 + x2 + 1)/(x2 - 1) dx
  1. Check Degrees: deg(x3 + x2 + 1) = 3, deg(x2 - 1) = 2. Since 3 ≥ 2, it's improper.
  2. Polynomial Long Division:
    (x3 + x2 + 1) / (x2 - 1) = (x + 1) + (x + 2)/(x2 - 1)
    (Quotient = x + 1, Remainder = x + 2)
  3. Partial Fractions for Remainder Term:
    (x + 2)/((x - 1)(x + 1)) = A/(x - 1) + B/(x + 1)
    Solving for A and B: A = 3/2, B = -1/2.
    So, (x + 2)/(x2 - 1) = (3/2)/(x - 1) - (1/2)/(x + 1)
  4. Integrate:
    ∫ (x + 1 + (3/2)/(x - 1) - (1/2)/(x + 1)) dx
    = x2/2 + x + (3/2)ln|x - 1| - (1/2)ln|x + 1| + C
💡 Prevention Tips:
  • Golden Rule: Make checking the degrees the *first* step for *every* rational function integration problem.
  • Master Long Division: Ensure you are proficient in polynomial long division. This is a foundational skill for JEE Advanced.
  • JEE Context: In JEE Advanced, questions involving improper rational functions are common traps. Always be vigilant!
JEE_Advanced
Important Other

❌ <span style='color: #FF0000;'>Incorrect Partial Fraction Decomposition Setup</span>

Students frequently err in setting up partial fractions, especially when dealing with repeated linear factors (e.g., (ax+b)n) or irreducible quadratic factors (e.g., (ax2+bx+c)). They might incorrectly assign a single term for repeated factors or a constant numerator for irreducible quadratics, leading to an invalid decomposition and incorrect integration.
💭 Why This Happens:
  • Lack of clarity on the specific rules for distinct linear, repeated linear, and irreducible quadratic factor types.
  • Confusion between these different forms.
  • Haste in setting up the decomposition, causing students to overlook crucial details.
✅ Correct Approach:
The correct setup for each type of factor in the denominator is critical for successful partial fraction decomposition:
  • For each distinct linear factor (ax+b), the term is: A/(ax+b)
  • For each repeated linear factor (ax+b)n, the terms are: A1/(ax+b) + A2/(ax+b)2 + ... + An/(ax+b)n
  • For each irreducible quadratic factor (ax2+bx+c), the term is: (Ax+B)/(ax2+bx+c)
📝 Examples:
❌ Wrong:

Consider the decomposition of 1 / [x(x-1)2(x2+4)]

Incorrect Setup:

1 / [x(x-1)2(x2+4)] = A/x + B/(x-1) + C/(x2+4)

Mistake: The repeated factor (x-1)2 is not fully accounted for (missing the (x-1)2 term), and the irreducible quadratic (x2+4) has an incorrect numerator form (should be linear, not a constant).

✅ Correct:

For the same expression 1 / [x(x-1)2(x2+4)]

Correct Setup:

1 / [x(x-1)2(x2+4)] = A/x + B/(x-1) + C/(x-1)2 + (Dx+E)/(x2+4)

Explanation:

  • A/x for the distinct linear factor 'x'.
  • B/(x-1) + C/(x-1)2 for the repeated linear factor '(x-1)2'.
  • (Dx+E)/(x2+4) for the irreducible quadratic factor '(x2+4)'.

💡 Prevention Tips:
  • Factorize Completely: Always ensure the denominator is fully factored into its simplest linear and irreducible quadratic terms.
  • Identify Factor Types: Clearly classify each factor (distinct linear, repeated linear, or irreducible quadratic) before setting up.
  • Master Forms: Thoroughly learn and accurately apply the correct partial fraction form for each identified factor type.
  • JEE Advanced Alert: While CBSE might focus on simpler cases, JEE Advanced often involves complex combinations of all factor types, requiring meticulous setup.
  • Check for Improper Fractions: Always perform polynomial long division if the numerator's degree is greater than or equal to the denominator's degree *before* attempting partial fraction decomposition.
JEE_Advanced
Important Calculation

❌ Algebraic Errors in Determining Partial Fraction Coefficients

A frequent calculation mistake in Integration by Partial Fractions involves errors in determining the unknown constants (A, B, C, etc.) after setting up the partial fraction decomposition. These are often algebraic slips, sign errors, or arithmetic errors during the process of equating coefficients or substituting specific values of 'x' to find the constants. An incorrect coefficient directly leads to a wrong final integral, even if the integration steps themselves are conceptually correct.
💭 Why This Happens:
  • Hurried Calculations: Students often rush through the algebraic steps to find A, B, C, leading to careless mistakes.
  • Sign Errors: Misplacing a negative sign during expansion or substitution is very common.
  • Arithmetic Mistakes: Simple addition, subtraction, or division errors while solving simultaneous equations or simplifying expressions.
  • Lack of Verification: Not cross-checking the determined coefficients by substituting them back into the original algebraic identity.
✅ Correct Approach:
Always adopt a systematic and careful approach. After setting up the partial fraction decomposition and forming the algebraic identity (e.g., numerator = A(factor1) + B(factor2)...), proceed methodically:
  • Method of Substitution: Substitute roots of the denominators (if distinct linear factors) one by one to find some constants directly.
  • Method of Equating Coefficients: For remaining constants, equate coefficients of like powers of 'x' on both sides of the identity to form simultaneous equations.
  • Verification: Before integrating, substitute the found values of A, B, C back into the algebraic identity and choose one more arbitrary value of 'x' (not a root) to ensure the equality holds. This simple check can catch most calculation errors.
📝 Examples:
❌ Wrong:

Consider the integral: ( int frac{3x+5}{(x-1)(x+2)} dx )
Decomposition: ( frac{3x+5}{(x-1)(x+2)} = frac{A}{x-1} + frac{B}{x+2} )
Algebraic identity: ( 3x+5 = A(x+2) + B(x-1) )
Wrong Calculation:
Substitute ( x=1 ): ( 3(1)+5 = A(1+2) + B(1-1) Rightarrow 8 = 3A Rightarrow A = 8/3 )
Substitute ( x=-2 ): ( 3(-2)+5 = A(-2+2) + B(-2-1) Rightarrow -6+5 = -3B Rightarrow -1 = -3B Rightarrow B = -1/3 )
Here, a sign error was made for B, making it negative instead of positive, leading to the wrong integral: ( int left( frac{8/3}{x-1} - frac{1/3}{x+2}
ight) dx = frac{8}{3}ln|x-1| - frac{1}{3}ln|x+2| + C )

✅ Correct:

For the same integral: ( int frac{3x+5}{(x-1)(x+2)} dx )
Algebraic identity: ( 3x+5 = A(x+2) + B(x-1) )
Correct Calculation:
Substitute ( x=1 ): ( 3(1)+5 = A(1+2) + B(1-1) Rightarrow 8 = 3A Rightarrow A = 8/3 )
Substitute ( x=-2 ): ( 3(-2)+5 = A(-2+2) + B(-2-1) Rightarrow -1 = -3B Rightarrow B = 1/3 )
Thus, the correct partial fraction decomposition is ( frac{8/3}{x-1} + frac{1/3}{x+2} )
The correct integral is: ( int left( frac{8/3}{x-1} + frac{1/3}{x+2}
ight) dx = frac{8}{3}ln|x-1| + frac{1}{3}ln|x+2| + C )

💡 Prevention Tips:
  • Double-Check Algebra: Carefully expand and combine like terms. Pay close attention to signs.
  • Systematic Substitution: Use the substitution method for roots first, as it often simplifies finding coefficients.
  • Equating Coefficients (JEE Advanced): For more complex denominators (e.g., irreducible quadratic factors, repeated factors), equating coefficients might be necessary. Be meticulous with setting up and solving simultaneous equations.
  • Verify Coefficients: Always substitute your calculated A, B, C values back into the original identity and test with one additional random 'x' value. This is a crucial step often skipped by students.
  • Practice: Regular practice with diverse problems builds precision and reduces calculation errors.
JEE_Advanced
Important Conceptual

❌ <span style='color: red;'>Ignoring Improper Rational Functions</span>

Students often forget that partial fraction decomposition is directly applicable only to proper rational functions (where the degree of the numerator is strictly less than the degree of the denominator). They attempt to decompose improper rational functions directly, leading to an incorrect setup and solution.
💭 Why This Happens:
This oversight stems from a lack of attention to the fundamental prerequisite of partial fraction decomposition. It can occur due to rushing through problems or having a weak conceptual foundation regarding the conditions for applying this powerful integration technique.
✅ Correct Approach:
Before applying partial fractions, always compare the degrees of the numerator, N(x), and the denominator, D(x).
📝 Examples:
❌ Wrong: 
∫ (x³ + x² - 1) / (x² - x) dx

Incorrect attempt:

Assume (x³ + x² - 1) / (x² - x) = A/x + B/(x-1)
✅ Correct: 
∫ (x³ + x² - 1) / (x² - x) dx

Correct approach:

1. Perform polynomial long division (since deg(N(x)) ≥ deg(D(x))):

   (x³ + x² - 1) / (x² - x) = (x + 2) + (2x - 1) / (x² - x)

2. Now, apply partial fractions only to the proper fraction:

   (2x - 1) / (x(x - 1)) = A/x + B/(x - 1)

3. The integral becomes: ∫(x + 2) dx + ∫(A/x + B/(x - 1)) dx
💡 Prevention Tips:
  • First Step Check: Always make comparing the degrees of the numerator and denominator your absolute first step when dealing with a rational function for integration.
  • Conceptual Clarity: Revisit the definition and conditions for applying partial fraction decomposition. Understanding *why* it works only for proper fractions is key.
  • Practice: Solve a variety of problems, specifically including those where polynomial long division is a mandatory prerequisite, to build this habit.
JEE_Advanced
Important Approximation

❌ Incorrect Partial Fraction Decomposition Setup

Students frequently make two crucial errors when applying partial fractions: 1) Neglecting to perform polynomial division for improper rational functions (where the degree of the numerator is greater than or equal to the degree of the denominator). 2) Employing an incorrect form for the partial fraction decomposition, particularly for repeated linear factors or irreducible quadratic factors. This foundational error in setting up the decomposition leads to incorrect coefficients and, consequently, an erroneous integral.
💭 Why This Happens:
This mistake primarily arises from a lack of a systematic approach, insufficient understanding of the prerequisites for partial fractions (like polynomial division), and not thoroughly grasping the specific decomposition forms required for different types of factors. Haste, focusing only on simpler cases, and rote memorization without conceptual understanding are major contributing factors.
✅ Correct Approach:
To avoid these errors, a precise, step-by-step approach is essential:

  • Check for Proper Fraction: Always compare the degree of the numerator with the degree of the denominator. If `deg(Numerator) ≥ deg(Denominator)`, perform polynomial long division first to express the rational function as `Polynomial + (Proper Fraction)`. Only the proper fraction part is then decomposed.

  • Apply Correct Forms: Understand and correctly apply the specific partial fraction forms for various denominator factors:

    • For a repeated linear factor `(ax+b)^n`: Use `A1/(ax+b) + A2/(ax+b)^2 + ... + An/(ax+b)^n`.

    • For an irreducible quadratic factor `(ax^2+bx+c)^n`: Use `(A1x+B1)/(ax^2+bx+c) + ... + (Anx+Bn)/(ax^2+bx+c)^n`.



📝 Examples:
❌ Wrong:
∫ (x^3 + 1) / (x^2 - 1) dx

Incorrect: Directly attempting to decompose `(x^3 + 1) / ((x-1)(x+1))` into `A/(x-1) + B/(x+1)`. This is wrong because the given fraction is improper (deg 3 / deg 2).


∫ 1 / (x(x+1)^2) dx

Incorrect: Decomposing as `A/x + B/(x+1)`. This misses the term required for the repeated factor `(x+1)^2`.
✅ Correct:
∫ (x^3 + 1) / (x^2 - 1) dx

Correct Steps:

  1. Polynomial Division: `(x^3 + 1) / (x^2 - 1) = x + (x + 1) / (x^2 - 1) = x + 1 / (x-1)` (for x≠-1).

  2. Integration: `∫ (x + 1/(x-1)) dx = x^2/2 + ln|x-1| + C`.




∫ 1 / (x(x+1)^2) dx

Correct Steps:

  1. Partial Fraction Form: `1 / (x(x+1)^2) = A/x + B/(x+1) + C/(x+1)^2`.

  2. Solving for A, B, C: (Detailed algebra omitted) You'd find `A=1, B=-1, C=-1`.

  3. Integration: `∫ (1/x - 1/(x+1) - 1/(x+1)^2) dx = ln|x| - ln|x+1| + 1/(x+1) + C`.

💡 Prevention Tips:

  • Always check degrees: Before proceeding, verify that the degree of the numerator is strictly less than the degree of the denominator. If not, perform polynomial long division.

  • Master all forms: Thoroughly understand and practice applying the correct partial fraction forms for every type of denominator factor: distinct linear, repeated linear, distinct irreducible quadratic, and repeated irreducible quadratic.

  • Follow a checklist: Adopt a systematic approach: (1) Is it improper? Divide. (2) Factor the denominator completely. (3) Set up the partial fraction decomposition using the correct forms. (4) Solve for the unknown coefficients. (5) Integrate each resulting term.

JEE_Main
Important Other

❌ Incorrect Partial Fraction Decomposition Setup

Students frequently make errors in setting up the partial fraction decomposition, especially when the denominator contains repeated linear factors or irreducible quadratic factors. This misapplication of rules leads to an incorrect breakdown of the rational function and, consequently, an incorrect integral.
💭 Why This Happens:
This mistake primarily stems from an insufficient understanding or recall of the specific rules for different types of factors in the denominator. Students often generalize the simplest case (distinct linear factors) to all scenarios, overlooking the nuances required for repeated or quadratic terms. Rushing the initial setup is a major contributing factor.
✅ Correct Approach:
Always analyze the denominator's factors carefully before setting up the partial fraction. Remember these specific rules for the denominator Q(x):
  • Distinct Linear Factor (ax+b): Term is A/(ax+b)
  • Repeated Linear Factor ((ax+b)n): Terms are A1/(ax+b) + A2/(ax+b)2 + ... + An/(ax+b)n
  • Irreducible Quadratic Factor (ax2+bx+c): Term is (Ax+B)/(ax2+bx+c)
  • Repeated Irreducible Quadratic Factor ((ax2+bx+c)n): Terms are (A1x+B1)/(ax2+bx+c) + ... + (Anx+Bn)/(ax2+bx+c)n
JEE Tip: Always ensure the rational function is proper (degree of numerator < degree of denominator) before decomposition. If not, perform polynomial long division first.
📝 Examples:
❌ Wrong:
For the integral ∫ x / ((x-1)2(x2+1)) dx,
Incorrect Setup: x / ((x-1)2(x2+1)) = A/(x-1) + (Bx+C)/(x2+1)
This misses the B/(x-1)2 term for the repeated linear factor.
✅ Correct:
For the integral ∫ x / ((x-1)2(x2+1)) dx,
Correct Setup: x / ((x-1)2(x2+1)) = A/(x-1) + B/((x-1)2) + (Cx+D)/(x2+1)
This correctly accounts for both the repeated linear factor and the irreducible quadratic factor.
💡 Prevention Tips:
  • Master the Forms: Thoroughly understand and memorize the four basic forms of partial fraction decomposition.
  • Categorize Factors: Before writing anything, mentally (or physically) identify each factor in the denominator as distinct linear, repeated linear, irreducible quadratic, or repeated irreducible quadratic.
  • Verify Proper Function: Always perform polynomial long division if the degree of the numerator is greater than or equal to the degree of the denominator.
  • Practice Variety: Solve problems involving all types of factors to build confidence and accuracy in the setup process.
JEE_Main
Important Sign Error

❌ Sign Error in Partial Fraction Decomposition

Students frequently make sign errors when determining the coefficients (A, B, C, etc.) in partial fraction decomposition, or when substituting these coefficients back into the partial fraction form. This typically happens during algebraic manipulation, especially when dealing with negative values for 'x' during substitution or while solving simultaneous equations.
💭 Why This Happens:
  • Carelessness: Rushing through calculations, leading to oversight of negative signs.
  • Algebraic Slips: Mistakes in distributing negative signs or combining terms incorrectly.
  • Substitution Errors: Incorrectly substituting a negative value of 'x' into an equation or mishandling the resulting negative product/sum.
  • Equation Solving: Errors while solving the system of linear equations for coefficients, particularly when one coefficient is negative.
✅ Correct Approach:
Always perform algebraic steps meticulously. Double-check each substitution and calculation involving negative numbers. When a coefficient is determined, ensure its sign is correctly carried forward into the partial fraction expression. For JEE Main, accuracy in these steps is crucial as a single sign error can lead to a completely wrong integral.
📝 Examples:
❌ Wrong:
Consider the integral of (frac{1}{(x+1)(x-2)}).
Decomposition: (frac{A}{x+1} + frac{B}{x-2})
(1 = A(x-2) + B(x+1))
Set (x=2): (1 = B(3) Rightarrow B = frac{1}{3})
Set (x=-1): (1 = A(-3) Rightarrow A = -frac{1}{3})
Wrong student step: Often, students might mistakenly write (A = frac{1}{3}) or in the next step, present the decomposition as (frac{1}{3(x+1)} + frac{1}{3(x-2)}) due to neglecting the negative sign of A. This would lead to the integral: (frac{1}{3}ln|x+1| + frac{1}{3}ln|x-2|).
✅ Correct:
For the same integral of (frac{1}{(x+1)(x-2)}):
Correct coefficients: (A = -frac{1}{3}) and (B = frac{1}{3})
The correct partial fraction decomposition is: (frac{-frac{1}{3}}{x+1} + frac{frac{1}{3}}{x-2})
Which can be written as: (frac{1}{3(x-2)} - frac{1}{3(x+1)})
Integrating this gives: (frac{1}{3}ln|x-2| - frac{1}{3}ln|x+1| + C).
This can also be written as (frac{1}{3}lnleft|frac{x-2}{x+1}
ight| + C).
💡 Prevention Tips:
  • Step-by-Step Verification: After finding each coefficient, quickly re-substitute it into the original equation to ensure it holds true, particularly checking the signs.
  • Highlight Negative Signs: When a negative sign appears, make a conscious effort to highlight or circle it to ensure it's not overlooked.
  • Use Parentheses: When substituting negative numbers, always use parentheses, e.g., (A(-3)) instead of (A-3), to prevent errors in multiplication or subtraction.
  • Compare Coefficients Method: If direct substitution is prone to errors for you, try the 'comparing coefficients' method as an alternative, but still be careful with signs in the system of equations.
  • Practice: Solve numerous problems to build confidence and reduce the likelihood of these common algebraic blunders.
JEE_Main
Important Other

❌ Ignoring Proper Fraction Condition

A common mistake is to directly apply partial fraction decomposition methods to improper rational functions. An improper rational function is one where the degree of the numerator is greater than or equal to the degree of the denominator (e.g., degree(N) ≥ degree(D)). The standard partial fraction decomposition formulas are strictly applicable only to proper rational functions (degree(N) < degree(D)). Failing to perform polynomial long division first for improper fractions leads to incorrect decomposition and integration.
💭 Why This Happens:
Students often overlook the fundamental prerequisite for partial fraction decomposition. This can stem from rushing through problems, not clearly understanding the definition of proper vs. improper rational functions, or simply memorizing the decomposition forms without grasping the underlying conditions. In CBSE, this crucial step is often a test of conceptual understanding.
✅ Correct Approach:
Before applying partial fraction decomposition, always check the degrees of the numerator and the denominator.

  • If degree(Numerator) < degree(Denominator), it is a proper fraction. Proceed directly with partial fraction decomposition.

  • If degree(Numerator) ≥ degree(Denominator), it is an improper fraction. First, perform polynomial long division to express the function in the form: Quotient + (Remainder / Denominator). The (Remainder / Denominator) part will now be a proper fraction, to which partial fraction decomposition can be applied.

📝 Examples:
❌ Wrong:
Consider the integral: ∫ (x² + 3x + 2) / (x + 1) dx

Wrong approach: Attempting to directly write (x² + 3x + 2) / (x + 1) = A / (x + 1) or similar partial fraction forms. This is incorrect because degree(Numerator) = 2 and degree(Denominator) = 1, making it an improper fraction.

✅ Correct:
For the integral: ∫ (x² + 3x + 2) / (x + 1) dx

Correct approach:



  1. Check degrees: degree(N)=2, degree(D)=1. It's an improper fraction.

  2. Perform polynomial long division:
    (x² + 3x + 2) / (x + 1) = x + 2 (since x² + 3x + 2 = (x+1)(x+2))

  3. Integrate:
    ∫ (x + 2) dx = x²/2 + 2x + C


JEE Note: Even if the division results in no remainder, as in this example, recognizing the improper fraction is key. For cases with remainder, the remainder term is then decomposed.

💡 Prevention Tips:

  • Always start by comparing degrees of numerator and denominator in rational functions.

  • Clearly identify if the given rational function is proper or improper.

  • Practice polynomial long division to become proficient in converting improper fractions.

  • For CBSE exams, explicitly showing the long division step when required can fetch marks, even if it seems trivial.

CBSE_12th
Important Approximation

❌ Incomplete Partial Fraction Decomposition for Repeated Linear Factors

Students frequently make an 'approximation' error by incorrectly setting up the partial fraction decomposition when the denominator contains repeated linear factors. Instead of including a separate term for each power of the repeated factor, they often miss out on intermediate terms, leading to an incorrect or incomplete algebraic form. This directly impacts the subsequent integration process.
💭 Why This Happens:
  • Lack of Conceptual Clarity: A superficial understanding of the rules governing partial fraction decomposition, especially for different types of denominator factors.
  • Haste and Overlooking Details: During exams, students might quickly identify a factor as repeated but fail to include all necessary terms in the decomposition.
  • Confusing with Simple Linear Factors: Mistaking a repeated linear factor like `(ax+b)^n` for a simple linear factor `(ax+b)` or `(ax+b)^n` as a single unit without its preceding powers.
✅ Correct Approach:
For a rational function where the denominator includes a repeated linear factor `(ax+b)^n` (where 'n' is an integer greater than 1), the partial fraction decomposition must include 'n' distinct terms, one for each power of `(ax+b)` from 1 up to 'n'.
For example, if the denominator has `(ax+b)^3`, the corresponding terms should be `A/(ax+b) + B/(ax+b)^2 + C/(ax+b)^3`.
📝 Examples:
❌ Wrong:
Consider the integral of `(5x-2)/((x-1)^2 * (x+3)) dx`.
A common incorrect partial fraction setup would be:
` (5x-2)/((x-1)^2 * (x+3)) = A/(x-1)^2 + B/(x+3) `
Here, the term `A/(x-1)` is missing, leading to an incorrect system for finding coefficients.
✅ Correct:
For the integral of `(5x-2)/((x-1)^2 * (x+3)) dx`.
The correct partial fraction decomposition must account for both powers of `(x-1)`:
` (5x-2)/((x-1)^2 * (x+3)) = A/(x-1) + B/(x-1)^2 + C/(x+3) `
This form allows for a unique solution for A, B, and C, ensuring correct integration. (This form is applicable for both CBSE and JEE.)
💡 Prevention Tips:
  • Strict Rule Adherence: Do not approximate the partial fraction form. Learn the exact rules for each type of factor (linear, repeated linear, irreducible quadratic, repeated irreducible quadratic).
  • Systematic Checklist: Before equating and solving for coefficients, always perform a quick check:
    • Are all factors in the denominator represented?
    • For repeated factors like `(ax+b)^n`, are all `n` terms (from power 1 to n) included?
  • Practice with Variety: Work through problems featuring diverse denominator structures, especially those with repeated factors, to solidify your understanding and avoid this 'approximation' trap.
CBSE_12th
Important Sign Error

❌ Incorrect Sign Application from Constants in Partial Fractions

Students often make sign errors when substituting the calculated values of constants (A, B, C, etc.) back into the partial fraction decomposition, or during the subsequent integration step, especially with negative constants. This leads to an incorrect sign in one or more terms of the final integrated expression.
💭 Why This Happens:
This mistake primarily stems from carelessness in algebraic manipulation, rushing through calculations, or misinterpreting the sign of a constant when writing out the decomposed fractions. Sometimes, students might correctly find A = -2, but then write it as +2/x in the decomposition. Another common source is sign errors while solving the system of linear equations for A, B, C.
✅ Correct Approach:
Always be meticulous with signs throughout the process.
  1. Solve for Constants Carefully: Double-check the system of equations used to find A, B, C.
  2. Substitute Correctly: When writing the partial fraction decomposition, ensure that each constant (A, B, C) is substituted with its exact sign. If A = -2, write it as -2/x, not +2/x.
  3. Integrate with Attention: During integration, pay close attention to the constant factor, including its sign. For example, ∫(-2/x) dx = -2log|x| + C, not 2log|x| + C.
📝 Examples:
❌ Wrong:
Consider the integral: ∫ (x-1) / (x(x+2)) dx
Decomposition: A/x + B/(x+2)
x-1 = A(x+2) + Bx
Setting x=0: -1 = 2A => A = -1/2
Setting x=-2: -3 = -2B => B = 3/2
Wrong decomposition: ∫ (1/2x + 3/(2(x+2))) dx (Mistake: A's sign is inverted)
Wrong Integration: (1/2)log|x| + (3/2)log|x+2| + C
✅ Correct:
Consider the integral: ∫ (x-1) / (x(x+2)) dx
Decomposition: A/x + B/(x+2)
x-1 = A(x+2) + Bx
Setting x=0: -1 = 2A => A = -1/2
Setting x=-2: -3 = -2B => B = 3/2
Correct decomposition: ∫ (-1/(2x) + 3/(2(x+2))) dx
Correct Integration: (-1/2)log|x| + (3/2)log|x+2| + C
💡 Prevention Tips:
  • Double-Check Algebraic Steps: Always re-verify the calculations for A, B, C.
  • Write Clearly: Ensure the sign of each constant is explicitly written when forming the partial fraction sum.
  • Use Parentheses: For clarity, especially with negative constants, you can write A/x as (-1/2)/x or -(1/(2x)).
  • Self-Correction: After finding the indefinite integral, differentiate it mentally or on scratch paper to see if you get the original integrand. This is a powerful check for both signs and magnitudes.

For CBSE Class 12 exams, even a single sign error can lead to significant loss of marks, so attention to detail is crucial!
CBSE_12th
Important Unit Conversion

❌ Ignoring Improper Rational Functions Before Partial Fraction Decomposition

A very common error students make is attempting to apply partial fraction decomposition directly to an improper rational function. An improper rational function is one where the degree of the numerator is greater than or equal to the degree of the denominator. Partial fraction decomposition is strictly applicable only to proper rational functions.
💭 Why This Happens:
This mistake often arises from a lack of thorough understanding of the prerequisites for applying partial fractions. Students might rush into the decomposition process without checking the fundamental condition, or they might confuse the conditions for polynomial division with those for partial fractions. For CBSE and JEE, this foundational step is crucial.
✅ Correct Approach:
Whenever you encounter a rational function for integration, the very first step is to check the degrees of the numerator and denominator.
  • If the function is proper (degree of numerator < degree of denominator), proceed directly with partial fraction decomposition.
  • If the function is improper (degree of numerator ≥ degree of denominator), you must first perform polynomial long division. This will transform the improper fraction into the sum of a polynomial and a proper rational function. Only then do you apply partial fraction decomposition to the proper rational function part.
📝 Examples:
❌ Wrong:
Consider ∫ (x³ + x + 1) / (x² - 1) dx.
Wrong: Directly assuming (x³ + x + 1) / (x² - 1) = A/(x-1) + B/(x+1). This form is incorrect because the degree of the numerator (3) is greater than the degree of the denominator (2).
✅ Correct:
For ∫ (x³ + x + 1) / (x² - 1) dx:
1. Perform polynomial long division:
(x³ + x + 1) ÷ (x² - 1) = x + (2x + 1) / (x² - 1)
So, ∫ (x³ + x + 1) / (x² - 1) dx = ∫ [x + (2x + 1) / (x² - 1)] dx
2. Now, apply partial fractions to the proper rational part (2x + 1) / (x² - 1).
(2x + 1) / [(x-1)(x+1)] = A/(x-1) + B/(x+1)
Solving for A and B gives A = 3/2, B = 1/2.
3. Integrate each part:
∫ [x + 3/(2(x-1)) + 1/(2(x+1))] dx = x²/2 + (3/2)ln|x-1| + (1/2)ln|x+1| + C
JEE Tip: Always look for simplification before partial fractions, sometimes factorizing the numerator or denominator can reduce complexity.
💡 Prevention Tips:
  • Always Check Degrees: Make it a habit to compare the degrees of the numerator and denominator as the first step for any rational function integration problem.
  • Master Polynomial Long Division: Ensure you are proficient in performing polynomial long division quickly and accurately.
  • Understand the 'Why': Remember that partial fractions work by decomposing a proper fraction into simpler proper fractions. An improper fraction needs to be 'broken down' into a polynomial (which integrates easily) and a proper fraction first.
CBSE_12th
Important Formula

❌ Incorrect Partial Fraction Decomposition for Repeated or Irreducible Quadratic Factors

Students frequently make errors in setting up the correct partial fraction decomposition, particularly when dealing with repeated linear factors or irreducible quadratic factors in the denominator. This misunderstanding directly impacts the subsequent integration steps.
  • For a repeated linear factor like (ax+b)², students might incorrectly write A/(ax+b), omitting the term for the higher power.
  • For an irreducible quadratic factor like (ax²+bx+c), they might incorrectly write A/(ax²+bx+c) instead of the required (Ax+B)/(ax²+bx+c).
This is a critical error as it leads to an entirely wrong setup for finding the constants and, consequently, an incorrect integral.
💭 Why This Happens:
This mistake stems from a lack of thorough understanding and memorization of the distinct rules for partial fraction decomposition based on the nature of the denominator's factors. Students often rush to decompose without carefully analyzing the factors' types (linear vs. quadratic, non-repeated vs. repeated). Confusion between these cases is common.
✅ Correct Approach:
The correct approach demands strict adherence to the partial fraction decomposition rules. For a rational function P(x)/Q(x) where the degree of P(x) < degree of Q(x):
  • Non-repeated linear factor (ax+b): Add a term A/(ax+b).
  • Repeated linear factor (ax+b)ⁿ: Add terms A₁/(ax+b) + A₂/(ax+b)² + ... + Aⁿ/(ax+b)ⁿ.
  • Non-repeated irreducible quadratic factor (ax²+bx+c): Add a term (Ax+B)/(ax²+bx+c).
  • Repeated irreducible quadratic factor (ax²+bx+c)ⁿ: Add terms (A₁x+B₁)/(ax²+bx+c) + (A₂x+B₂)/(ax²+bx+c)² + ... + (Aⁿx+Bⁿ)/(ax²+bx+c)ⁿ.
📝 Examples:
❌ Wrong:
For the integral ∫(x+1)/(x(x-2)²) dx, a common incorrect decomposition is:
(x+1)/(x(x-2)²) = A/x + B/(x-2) (Missing the C/(x-2)² term).
✅ Correct:
For the integral ∫(x+1)/(x(x-2)²) dx, the correct decomposition, recognizing the repeated linear factor (x-2)², is:
(x+1)/(x(x-2)²) = A/x + B/(x-2) + C/(x-2)²
Similarly, for ∫(2x)/(x²+1) dx, the decomposition would be (2x)/(x²+1), which doesn't need partial fractions if solved directly, but if part of a larger denominator, e.g., ∫(2x)/((x²+1)(x-1)) dx, the quadratic factor term is (Ax+B)/(x²+1).
💡 Prevention Tips:
  • Master the Four Cases: Dedicate time to thoroughly understand and memorize the specific decomposition formats for all types of factors. Practice recognizing them instantly.
  • Systematic Factorization: Always start by completely factoring the denominator. If a quadratic factor cannot be factored further over real numbers, it's irreducible.
  • Verify Denominator Degree: Before decomposition, ensure the degree of the numerator is strictly less than the degree of the denominator. If not, perform polynomial long division first (this is a common step missed in CBSE exams).
  • Check Your Setup: Before proceeding to find the constants, double-check that your partial fraction setup correctly accounts for every factor type and its multiplicity.
CBSE_12th
Important Calculation

❌ Algebraic Errors in Determining Coefficients (A, B, C)

Students frequently make algebraic blunders while solving for the unknown constants (A, B, C, etc.) after correctly setting up the partial fraction decomposition. These errors can occur during substitution of specific x-values or when comparing coefficients of like powers of x, leading to incorrect values for the constants and subsequently, an incorrect final integral.
💭 Why This Happens:
  • Carelessness in Substitution: Minor sign errors or arithmetic mistakes when substituting values of x.
  • Errors in Expansion and Collection: Mistakes when expanding terms and collecting like powers of x before comparing coefficients.
  • Sign Errors: Mismanaging negative signs during multiplication, division, or transposition of terms.
  • Lack of Systematic Approach: Not performing steps meticulously, especially when dealing with multiple constants or complex expressions.
✅ Correct Approach:
To accurately determine the coefficients:
  • Strategic Substitution: Substitute the roots of the denominator's factors (e.g., if a factor is (x-a), substitute x=a) to quickly find some constants.
  • Coefficient Comparison: For factors that don't have distinct roots, or to find remaining constants, expand the expression and compare the coefficients of corresponding powers of x on both sides of the equation.
  • Meticulous Algebra: Pay extreme attention to signs, parentheses, and basic arithmetic operations at every step.
  • Cross-Verification: If time permits, substitute a random value of x (not used for finding roots) into the original and decomposed equation to check if the constants are correct.
📝 Examples:
❌ Wrong:
Consider the integral of (3x + 1) / ((x-1)(x+2)) dx.
Correct decomposition setup: (3x + 1) / ((x-1)(x+2)) = A/(x-1) + B/(x+2)
Therefore, 3x + 1 = A(x+2) + B(x-1)
To find A, let x = 1:
3(1) + 1 = A(1+2) + B(1-1)
4 = 3A
Common Mistake: A student might incorrectly write A = 4/3, but then make a sign error when solving for B.
To find B, let x = -2:
3(-2) + 1 = A(-2+2) + B(-2-1)
-6 + 1 = 0 + B(-3)
-5 = -3B
Mistake: Student incorrectly calculates B = -5/3 instead of the correct B = 5/3.
✅ Correct:
Following the same problem:
3x + 1 = A(x+2) + B(x-1)
Step 1: Find A by substituting x = 1
3(1) + 1 = A(1+2) + B(1-1)
4 = 3A
A = 4/3
Step 2: Find B by substituting x = -2
3(-2) + 1 = A(-2+2) + B(-2-1)
-6 + 1 = 0 + B(-3)
-5 = -3B
B = 5/3
The correct decomposition is then (4/3)/(x-1) + (5/3)/(x+2).
💡 Prevention Tips:
  • Work Step-by-Step: Avoid rushing through algebraic calculations. Write down each step clearly.
  • Double-Check Signs: This is crucial. A single sign error can invalidate the entire solution.
  • Systematic Approach: Use a consistent method for finding coefficients (e.g., always substitute roots first, then compare coefficients if needed).
  • Practice Regularly: Solve a variety of problems to build proficiency and identify common pitfalls. For CBSE exams, ensure your algebraic steps are neat and easy to follow for partial credit.
CBSE_12th
Important Conceptual

❌ Improper Partial Fraction Decomposition Setup

Students frequently make errors in setting up the partial fraction decomposition. This includes:
  • Failing to perform polynomial long division when the numerator's degree is greater than or equal to the denominator's.
  • Incorrectly handling repeated linear factors (e.g., $(ax+b)^2$) or irreducible quadratic factors (e.g., $(ax^2+bx+c)$) in the decomposition form.
💭 Why This Happens:
This mistake stems from a lack of thorough understanding of the conditions for applying partial fractions (specifically, that the rational function must be proper) and the precise decomposition rules for different types of denominator factors (linear vs. quadratic, distinct vs. repeated).
✅ Correct Approach:
  1. Degree Check: Always compare the degree of the numerator (N) with the degree of the denominator (D). If deg(N) ≥ deg(D), perform polynomial long division first: $frac{N(x)}{D(x)} = Q(x) + frac{R(x)}{D(x)}$, where deg(R) < deg(D). Then, apply partial fractions only to $frac{R(x)}{D(x)}$.
  2. Factorization: Fully factorize the denominator into linear and irreducible quadratic factors.
  3. Decomposition Rules: Apply the correct partial fraction forms:
    • For a distinct linear factor $(ax+b)$: $frac{A}{ax+b}$
    • For a repeated linear factor $(ax+b)^n$: $frac{A_1}{ax+b} + frac{A_2}{(ax+b)^2} + dots + frac{A_n}{(ax+b)^n}$
    • For a distinct irreducible quadratic factor $(ax^2+bx+c)$: $frac{Ax+B}{ax^2+bx+c}$
📝 Examples:
❌ Wrong:

For $int frac{x^2}{x^2-1} dx$, a common mistake is to directly assume $frac{x^2}{(x-1)(x+1)} = frac{A}{x-1} + frac{B}{x+1}$. This is incorrect because the degree of the numerator equals the degree of the denominator.

✅ Correct:

For $int frac{x^2}{x^2-1} dx$:

  1. Degree Check: The degree of the numerator (2) is equal to the degree of the denominator (2). Hence, perform long division: $frac{x^2}{x^2-1} = frac{(x^2-1)+1}{x^2-1} = 1 + frac{1}{x^2-1}$.
  2. Partial Fraction Decomposition: Now, decompose the proper fraction $frac{1}{x^2-1}$.$frac{1}{(x-1)(x+1)} = frac{A}{x-1} + frac{B}{x+1}$.Solving for A and B yields $A=1/2$ and $B=-1/2$.
  3. Integration: Integrate the resulting terms:$int left(1 + frac{1/2}{x-1} - frac{1/2}{x+1}
    ight) dx = x + frac{1}{2}ln|x-1| - frac{1}{2}ln|x+1| + C = x + frac{1}{2}lnleft|frac{x-1}{x+1}
    ight| + C$.
💡 Prevention Tips:
  • Always begin with a degree check. This is a critical first step for any rational function integration problem using partial fractions. (Applicable for both CBSE and JEE).
  • Memorize and understand the four basic partial fraction forms. Practice identifying which form to use based on the denominator's factors.
  • Carefully factorize the denominator. A mistake here will lead to an incorrect setup.
CBSE_12th
Important Conceptual

❌ Ignoring Polynomial Long Division Before Partial Fractions

Students often directly apply partial fraction decomposition rules to rational functions where the degree of the numerator polynomial (N(x)) is greater than or equal to the degree of the denominator polynomial (D(x)). This is a fundamental conceptual error, as partial fraction decomposition is strictly applicable only to proper rational functions (where deg(N(x)) < deg(D(x))).
💭 Why This Happens:
This mistake primarily occurs due to:
  • Lack of understanding of the prerequisite condition for partial fraction decomposition.
  • Rushing to apply decomposition rules without first checking the degrees of the polynomials.
  • Forgetting the concept of improper rational functions and their transformation via long division.
✅ Correct Approach:
Before attempting partial fraction decomposition, always compare the degrees of the numerator and denominator:
  • If deg(N(x)) < deg(D(x)), proceed directly with partial fraction decomposition.
  • If deg(N(x)) ≥ deg(D(x)), first perform polynomial long division to express the rational function as:
    N(x)/D(x) = Q(x) + R(x)/D(x)
    where Q(x) is the quotient polynomial and R(x)/D(x) is a proper rational function (i.e., deg(R(x)) < deg(D(x))). Then, apply partial fraction decomposition only to the R(x)/D(x) term.
📝 Examples:
❌ Wrong:
Trying to directly decompose ∫ (x² + 1) / (x - 1) dx into A/(x-1) without first performing polynomial long division. This is incorrect because deg(x² + 1) = 2 and deg(x - 1) = 1 (i.e., deg(N(x)) ≥ deg(D(x))).
✅ Correct:
For ∫ (x² + 1) / (x - 1) dx:
  • First, perform long division: (x² + 1) / (x - 1) = x + 1 + 2 / (x - 1).
  • Then integrate: ∫ (x + 1 + 2 / (x - 1)) dx = x²/2 + x + 2 ln|x - 1| + C. (Partial fractions are applied to the remainder term, R(x)/D(x), if it is a more complex proper rational function).
💡 Prevention Tips:
  • Always check degrees: Make it a habit to compare deg(N(x)) and deg(D(x)) as the absolute first step for any rational function integration problem.
  • Master Polynomial Long Division: Ensure you are proficient in polynomial long division, as it's a fundamental prerequisite.
  • JEE Main Tip: Many JEE problems subtly include this trap. Always be vigilant!
JEE_Main
Important Calculation

❌ Omitting Polynomial Division for Improper Fractions

Students frequently apply partial fraction decomposition methods directly when the numerator's degree is greater than or equal to the denominator's degree. This is a critical algebraic error, as partial fractions are exclusively for proper rational functions.
💭 Why This Happens:
This common mistake stems from a fundamental oversight: partial fraction decomposition is only valid for proper rational functions (where degree(Numerator) < degree(Denominator)). Students often skip this initial degree check, assuming direct applicability.
✅ Correct Approach:
For any rational function P(x)/Q(x), always first compare their degrees. If degree(P(x)) ≥ degree(Q(x)), you must perform polynomial long division to get: P(x)/Q(x) = Quotient(x) + Remainder(x)/Q(x). Partial fractions are then applied only to the Remainder(x)/Q(x) term, which is now a proper rational function. This step is vital for JEE Main.
📝 Examples:
❌ Wrong:

Problem: Integrate ∫ (x³ + 1) / (x² - 4) dx

Incorrect: Directly setting (x³ + 1) / ((x-2)(x+2)) = A/(x-2) + B/(x+2). This skips the prerequisite polynomial division.

✅ Correct:

Problem: Integrate ∫ (x³ + 1) / (x² - 4) dx

  1. Polynomial Division: (x³ + 1) / (x² - 4) = x + (4x + 1) / (x² - 4).
  2. Partial Fractions: Decompose (4x + 1) / ((x-2)(x+2)) into 9/(4(x-2)) + 7/(4(x+2)).
  3. Integration: x²/2 + (9/4)ln|x-2| + (7/4)ln|x+2| + C.
💡 Prevention Tips:
  • Initial Check: Always compare degrees (Numerator vs. Denominator) first.
  • Master Division: Ensure proficiency in polynomial long division.
  • Conceptual Clarity: Partial fractions are exclusively for proper rational functions.
JEE_Main
Important Formula

❌ Incorrect Partial Fraction Decomposition for Repeated Linear Factors

Students often apply the partial fraction formula incorrectly when the denominator contains repeated linear factors. Instead of including terms for each power of the repeated factor up to its multiplicity, they might only include a single term or an incorrect set of terms.
💭 Why This Happens:
This mistake stems from a lack of complete understanding of the partial fraction decomposition rules. Students might:
  • Confuse repeated linear factors with distinct linear factors.
  • Rely on rote memorization without grasping the underlying principle of ensuring all possible lower degree polynomials are represented in the numerator.
  • Rush through the decomposition step, which is critical for successful integration.
✅ Correct Approach:
For a repeated linear factor (ax+b)n in the denominator, the partial fraction decomposition must include 'n' terms, with increasing powers of the factor in the denominator, each with a constant numerator. Specifically, for (ax+b)n, the terms should be:
$$ frac{A_1}{ax+b} + frac{A_2}{(ax+b)^2} + dots + frac{A_n}{(ax+b)^n} $$
📝 Examples:
❌ Wrong:

Consider the integral of $$ frac{x}{(x-1)^2 (x+2)} dx $$
A common incorrect decomposition is:

$$ frac{A}{(x-1)^2} + frac{B}{x+2} $$

This misses a crucial term for the first power of the repeated factor.

✅ Correct:

The correct partial fraction decomposition for the above expression is:

$$ frac{A}{x-1} + frac{B}{(x-1)^2} + frac{C}{x+2} $$

Here, the repeated factor (x-1)2 correctly gives rise to two terms: A/(x-1) and B/(x-1)2.

💡 Prevention Tips:
  • Understand the Rules: Do not just memorize formulas; understand why each term is necessary in the decomposition.
  • Systematic Approach: For each factor in the denominator, follow the specific rule: distinct linear, repeated linear, distinct irreducible quadratic, repeated irreducible quadratic.
  • Practice Regularly: Solve a variety of problems involving different types of factors to solidify your understanding.
  • Verify Degree: Ensure that the degree of the numerator in the original rational function is strictly less than the degree of the denominator. If not, perform polynomial long division first (JEE Main specific).
JEE_Main
Important Unit Conversion

❌ Ignoring Proper Fraction Requirement Before Partial Fraction Decomposition

A common and critical error is attempting to decompose an improper rational function into partial fractions directly, without first performing polynomial long division. Partial fraction decomposition is strictly applicable only to proper rational functions (where the degree of the numerator is less than the degree of the denominator).
💭 Why This Happens:
Students often jump straight to setting up the partial fraction form without checking the degrees of the numerator and denominator. This oversight stems from a lack of foundational understanding of the conditions required for partial fraction decomposition or simply rushing through the problem.
✅ Correct Approach:
Before attempting partial fraction decomposition, always compare the degrees of the numerator and denominator:
  • If degree(Numerator) < degree(Denominator): The function is a proper rational function. Proceed directly with partial fraction decomposition.
  • If degree(Numerator) ≥ degree(Denominator): The function is an improper rational function. First, perform polynomial long division to express the function as the sum of a polynomial and a proper rational function. Then, apply partial fraction decomposition only to the resulting proper rational function part.
📝 Examples:
❌ Wrong:

Consider the integral: ( int frac{x^3 + 1}{x^2 - x} dx )
Incorrect approach: Directly setting up partial fractions like ( frac{x^3 + 1}{x(x-1)} = frac{A}{x} + frac{B}{x-1} ) because the degree of the numerator (3) is greater than the degree of the denominator (2).

✅ Correct:

For ( int frac{x^3 + 1}{x^2 - x} dx ):
1. Observe that degree(Numerator) = 3 and degree(Denominator) = 2. It's an improper fraction.
2. Perform polynomial long division:

      x + 1

x^2-x | x^3     + 1

        -(x^3 - x^2)

        ----------

              x^2 + 1

            -(x^2 - x)

            ----------

                  x + 1

So, ( frac{x^3 + 1}{x^2 - x} = x + 1 + frac{x+1}{x^2-x} )
3. Now, apply partial fraction decomposition to the proper fraction ( frac{x+1}{x^2-x} ) (which is ( frac{x+1}{x(x-1)} )):
( frac{x+1}{x(x-1)} = frac{A}{x} + frac{B}{x-1} )
Solving for A and B gives A = -1 and B = 2.
4. The integral becomes ( int left( x + 1 + frac{-1}{x} + frac{2}{x-1}
ight) dx )
( = frac{x^2}{2} + x - ln|x| + 2ln|x-1| + C )

💡 Prevention Tips:
  • Always Check Degrees: Make it a habit to compare the degrees of the numerator and denominator as the very first step in any partial fraction problem.
  • Master Polynomial Long Division: Ensure you are proficient in polynomial long division. This is a prerequisite skill for handling improper rational functions.
  • JEE Specific: In JEE Mains, questions might deliberately present improper fractions to test this foundational understanding. Ignoring this step will lead to incorrect decomposition and, consequently, an incorrect final integral.
JEE_Main
Critical Unit Conversion

❌ Incorrect Partial Fraction Decomposition Setup (Irrelevance of Unit Conversion)

A critical mistake in 'Integration by Partial Fractions' is incorrectly setting up the partial fraction decomposition. Students often misapply the rules for repeated factors or irreducible quadratic factors. Crucially, some students might erroneously try to introduce or worry about 'unit conversion' in this context. It's vital to understand that Integration by Partial Fractions is a purely algebraic technique; unit conversion concepts (relevant in physics or applied problems) are entirely inapplicable here and reflect a fundamental misunderstanding of the topic's nature.
💭 Why This Happens:
This error stems from a lack of clear understanding regarding the distinct rules for different types of factors in the denominator (e.g., distinct linear, repeated linear, irreducible quadratic). The confusion about unit conversion often arises from rote learning or misinterpreting the 'problem-solving context' where mathematical tools might be applied, without grasping the boundaries of pure mathematical techniques.
✅ Correct Approach:
The correct approach involves systematically identifying the type of factors in the denominator and applying the corresponding partial fraction form. Unit conversion is never involved in this process.
  • For distinct linear factors (e.g., (ax+b)(cx+d)), the form is A/(ax+b) + B/(cx+d).
  • For repeated linear factors (e.g., (ax+b)n), the form is A₁/(ax+b) + A₂/(ax+b)² + ... + An/(ax+b)ⁿ.
  • For irreducible quadratic factors (e.g., (ax²+bx+c)), the form is (Ax+B)/(ax²+bx+c).
📝 Examples:
❌ Wrong:
Consider the integral of ∫ (x / ((x-1)(x-2)²)) dx.
Wrong Setup:
x / ((x-1)(x-2)²) = A/(x-1) + B/(x-2)  // Incorrect: Fails to account for (x-2)² fully.

Any attempt to convert 'x' or 'dx' into different units.
✅ Correct:
For the integral of ∫ (x / ((x-1)(x-2)²)) dx.
Correct Setup:
x / ((x-1)(x-2)²) = A/(x-1) + B/(x-2) + C/(x-2)²  // Correct: Accounts for distinct and repeated linear factors.
💡 Prevention Tips:
  • Master Decomposition Rules: Thoroughly learn and practice the specific partial fraction forms for distinct linear, repeated linear, and irreducible quadratic factors.
  • Check for Proper Fractions: Always ensure the degree of the numerator is less than the degree of the denominator. If not, perform polynomial long division first.
  • Conceptual Clarity: Understand that 'Integration by Partial Fractions' is a tool for algebraic manipulation to simplify integration. It is purely mathematical, and concepts like 'unit conversion' are outside its scope.
  • CBSE vs. JEE: Both exams expect precise application of these decomposition rules. JEE may involve more complex algebraic expressions, but the fundamental setup rules remain the same.
CBSE_12th
Critical Approximation

❌ <span style='color: #FF0000;'>Ignoring Improper Fractions: No Polynomial Long Division</span>

A critical mistake is attempting to apply partial fraction decomposition directly to improper rational functions. An improper rational function is one where the degree of the numerator is greater than or equal to the degree of the denominator (i.e., deg(N) ≥ deg(D)). Students often overlook this prerequisite, leading to incorrect decomposition and integration.
💭 Why This Happens:
This error stems from a fundamental misunderstanding of the conditions for partial fraction decomposition. The technique is strictly applicable only to proper rational functions (deg(N) < deg(D)). Students either forget this initial check or don't recognize an improper fraction, treating it as if it were proper, which is a form of 'approximating' the fraction's structure incorrectly.
✅ Correct Approach:
Before applying partial fraction decomposition, always compare the degrees of the numerator and denominator. If deg(N) ≥ deg(D), perform polynomial long division first. This will express the improper fraction as a sum of a polynomial and a proper rational function. Only the proper rational function part is then decomposed using partial fractions.
📝 Examples:
❌ Wrong:
Consider $int frac{x^3}{x^2 - 4} dx$.
Wrong approach: Directly assuming $frac{x^3}{x^2 - 4} = frac{A}{x-2} + frac{B}{x+2}$. This is incorrect because deg($x^3$) = 3 and deg($x^2 - 4$) = 2. Since 3 ≥ 2, it's an improper fraction.
✅ Correct:
For $int frac{x^3}{x^2 - 4} dx$:
1. Perform polynomial long division: $frac{x^3}{x^2 - 4} = x + frac{4x}{x^2 - 4}$.
2. Now, decompose only the proper fraction $frac{4x}{x^2 - 4}$ using partial fractions:
$frac{4x}{x^2 - 4} = frac{4x}{(x-2)(x+2)} = frac{A}{x-2} + frac{B}{x+2}$.
Solving for A and B gives A=2, B=2.
3. So, $frac{x^3}{x^2 - 4} = x + frac{2}{x-2} + frac{2}{x+2}$.
4. Integrate term by term: $int (x + frac{2}{x-2} + frac{2}{x+2}) dx = frac{x^2}{2} + 2 ln|x-2| + 2 ln|x+2| + C$.
💡 Prevention Tips:
  • Always Check Degrees: Make it a habit to compare the degrees of the numerator and denominator as the very first step in any partial fraction problem.
  • Polynomial Long Division Mastery: Ensure you are proficient in polynomial long division, as it's a critical prerequisite for many integration problems.
  • JEE & CBSE Alert: This step is fundamental and tested thoroughly in both CBSE board exams and JEE. Skipping it guarantees zero marks for the decomposition part.
CBSE_12th
Critical Other

❌ Ignoring Polynomial Long Division Before Partial Fractions

A critical mistake students make is attempting to apply partial fraction decomposition directly when the degree of the numerator polynomial (N(x)) is greater than or equal to the degree of the denominator polynomial (D(x)). This is a fundamental misunderstanding of the prerequisites for partial fraction decomposition.
💭 Why This Happens:
This error stems from a lack of thorough understanding of the conditions under which partial fractions can be applied. Students often jump straight to factorization and decomposition without first checking the relative degrees of the polynomials, perhaps due to time pressure or overlooking this initial, crucial step. It's a common oversight in both CBSE and JEE exams.
✅ Correct Approach:
When `deg(N(x)) ≥ deg(D(x))`, you must first perform polynomial long division. This transforms the improper rational function `N(x)/D(x)` into the form `Q(x) + R(x)/D(x)`, where `Q(x)` is the quotient polynomial and `R(x)` is the remainder polynomial, with the essential condition that `deg(R(x)) < deg(D(x))`. Only then can you apply partial fraction decomposition to the proper rational function `R(x)/D(x)`.
📝 Examples:
❌ Wrong:
Consider ∫(x³ + x) / (x² - 4) dx.
Incorrect Approach: Trying to directly write `(x³ + x) / ((x-2)(x+2)) = A/(x-2) + B/(x+2)` (This is algebraically incorrect as the LHS is an improper fraction).
✅ Correct:
For ∫(x³ + x) / (x² - 4) dx:
  1. Perform Polynomial Long Division:
    `(x³ + x) / (x² - 4) = x + (5x) / (x² - 4)`
  2. Decompose the proper fraction:
    `5x / (x² - 4) = 5x / ((x-2)(x+2))`
    `5x / ((x-2)(x+2)) = A/(x-2) + B/(x+2)`
    Solving for A and B gives `A = 5/2`, `B = 5/2`.
  3. Integrate:
    ∫ `(x³ + x) / (x² - 4) dx` = ∫ `x dx` + ∫ `(5/2)/(x-2) dx` + ∫ `(5/2)/(x+2) dx`
    = `x²/2 + (5/2)ln|x-2| + (5/2)ln|x+2| + C`
💡 Prevention Tips:
  • First Check: Always compare the degrees of the numerator and denominator before proceeding with partial fractions.
  • If Improper: If `deg(Numerator) ≥ deg(Denominator)`, perform polynomial long division immediately. This is a non-negotiable step.
  • Separate Integration: Remember to integrate the quotient polynomial `Q(x)` obtained from long division, along with the terms from the partial fraction decomposition of the remainder.
CBSE_12th
Critical Formula

❌ Incorrect Application of Partial Fraction Decomposition Forms

A critical mistake students make is incorrectly applying the partial fraction decomposition form based on the type of factors in the denominator. This often occurs with repeated linear factors or irreducible quadratic factors, where the assumed form for the partial fraction is incomplete or incorrect, leading to an unsolvable system for constants and ultimately wrong integration.
💭 Why This Happens:
This error stems from a lack of precise understanding of the rules governing partial fraction decomposition for different types of denominator factors. Students might rush, misremember, or oversimplify the required form, especially under exam pressure. The fundamental 'formula' for each factor type is either not known or confused.
✅ Correct Approach:
Always meticulously identify the type of each factor in the denominator and apply its corresponding partial fraction decomposition rule.
  • Distinct Linear Factor (ax+b): Use A/(ax+b)
  • Repeated Linear Factor (ax+b)n: Use A1/(ax+b) + A2/(ax+b)2 + ... + An/(ax+b)n
  • Irreducible Quadratic Factor (ax2+bx+c): Use (Ax+B)/(ax2+bx+c)

Important for CBSE & JEE: Ensure the fraction is proper. If improper, perform polynomial long division first.
📝 Examples:
❌ Wrong:
For the integral ∫ (x+5) / [(x-2)2 * (x2+1)] dx, a common mistake in *formula understanding* is to attempt the partial fraction decomposition as:
(x+5) / [(x-2)2 * (x2+1)] = A/(x-2)2 + (Bx+C)/(x2+1)
This setup is incorrect because it misses the term for the distinct linear factor `(x-2)` which is part of the repeated factor `(x-2)^2`.
✅ Correct:
The correct partial fraction decomposition for the integrand (x+5) / [(x-2)2 * (x2+1)] should correctly account for all factor types:
(x+5) / [(x-2)2 * (x2+1)] = A/(x-2) + B/(x-2)2 + (Cx+D)/(x2+1)
Here, A/(x-2) + B/(x-2)2 correctly represents the repeated linear factor `(x-2)^2`, and (Cx+D)/(x2+1) represents the irreducible quadratic factor `(x2+1)`.
💡 Prevention Tips:
  • Master the Forms: Spend dedicated time to memorize and understand the logic behind each partial fraction decomposition form.
  • Factor Analysis: Before decomposition, explicitly write down the type of each factor (linear, repeated linear, irreducible quadratic).
  • Check for Completeness: Always cross-check that every power of a repeated factor has a corresponding term and that irreducible quadratics have linear numerators.
  • Practice Diligently: Solve a variety of problems involving all combinations of factor types to solidify your understanding.
CBSE_12th
Critical Conceptual

❌ Neglecting polynomial long division for improper rational functions

Students frequently attempt to apply partial fraction decomposition directly to rational functions where the degree of the numerator is greater than or equal to the degree of the denominator (known as improper rational functions). This is a fundamental conceptual error, as partial fraction decomposition rules are primarily designed for proper rational functions only.
💭 Why This Happens:
This mistake stems from a lack of understanding of the prerequisite condition for partial fraction decomposition. Students often memorize the decomposition forms for various factor types without recognizing the initial step: ensuring the rational function is proper. They might rush to apply the templates without verifying the basic algebraic requirement.
✅ Correct Approach:
Before initiating any partial fraction decomposition, always compare the degrees of the numerator and denominator. If the degree of the numerator is greater than or equal to the degree of the denominator, you must first perform polynomial long division. This transforms the improper rational function into a sum of a polynomial and a proper rational function. Only the resulting proper rational function can then be decomposed using partial fractions.
📝 Examples:
❌ Wrong:
Consider $int frac{x^3 + x}{x^2-1} dx$.
Wrong approach: Directly attempting decomposition as $frac{x^3+x}{x^2-1} = frac{A}{x-1} + frac{B}{x+1}$. This is incorrect because the degree of the numerator (3) is greater than the degree of the denominator (2). Solving for A and B from this setup will not yield valid coefficients, as the form does not account for the polynomial part.
✅ Correct:
Consider $int frac{x^3 + x}{x^2-1} dx$.
Correct approach:
1. Check degrees: Degree(Numerator) = 3, Degree(Denominator) = 2. Since 3 $geq$ 2, it's an improper fraction.
2. Perform polynomial long division:
$frac{x^3 + x}{x^2-1} = x + frac{2x}{x^2-1}$ (Quotient is $x$, Remainder is $2x$)
3. Now, decompose the proper rational part, $frac{2x}{x^2-1}$, using partial fractions:
$frac{2x}{(x-1)(x+1)} = frac{A}{x-1} + frac{B}{x+1}$
Multiplying by $(x-1)(x+1)$: $2x = A(x+1) + B(x-1)$.
For $x=1$, $2(1) = A(1+1) Rightarrow A = 1$.
For $x=-1$, $2(-1) = B(-1-1) Rightarrow B = 1$.
4. Substitute back and integrate:
$int frac{x^3 + x}{x^2-1} dx = int (x + frac{1}{x-1} + frac{1}{x+1}) dx$
$= frac{x^2}{2} + ln|x-1| + ln|x+1| + C$
💡 Prevention Tips:
  • Systematic Check: Always begin by comparing the degrees of the numerator and denominator. Make this your first step for every partial fraction problem.
  • Master Long Division: Practice polynomial long division thoroughly. It is a prerequisite skill for handling improper rational functions.
  • Conceptual Clarity (JEE Focus): Understand why partial fractions work for proper functions. This deeper understanding will prevent common errors. For CBSE, direct application of the rule after division is sufficient.
CBSE_12th
Critical Calculation

❌ Calculation Errors in Determining Coefficients (A, B, C...) in Partial Fractions

Students frequently make arithmetic and algebraic errors while solving for the unknown coefficients (A, B, C, etc.) after setting up the partial fraction decomposition. This is a critical calculation mistake as an incorrect coefficient renders the entire subsequent integration wrong, leading to zero marks for the entire problem even if the setup and final integration steps are conceptually correct.
💭 Why This Happens:
  • Arithmetic Mistakes: Simple errors in addition, subtraction, multiplication, or division while equating coefficients or substituting values of 'x'.
  • Sign Errors: Incorrectly handling negative signs during algebraic manipulation.
  • Simultaneous Equation Errors: Mistakes when solving systems of linear equations (especially for three or more variables) to find the coefficients.
  • Carelessness: Rushing through calculations without double-checking.
✅ Correct Approach:
After setting up the partial fraction decomposition, use either the substitution method (substituting roots of the denominator) or the equating coefficients method, or a combination of both. Always cross-verify your calculations. For CBSE, showing clear steps for finding coefficients is crucial. For JEE, speed and accuracy are paramount; sometimes, a quick mental check or re-substitution can save marks.
📝 Examples:
❌ Wrong:
Consider the decomposition:
( frac{3x+5}{(x-1)(x+2)} = frac{A}{x-1} + frac{B}{x+2} )
After multiplying by ((x-1)(x+2)), we get ( 3x+5 = A(x+2) + B(x-1) ).
If a student substitutes (x=1), they might incorrectly calculate:
3(1)+5 = A(1+2) + B(1-1)
8 = 2A Rightarrow A = 4 (Mistake: 1+2 = 3, not 2).
This single error makes 'A' incorrect, leading to a wrong final answer.
✅ Correct:
Following the setup: ( 3x+5 = A(x+2) + B(x-1) ).
To find A: Substitute (x=1) (root of (x-1)).
3(1)+5 = A(1+2) + B(1-1)
8 = A(3) + 0
A = frac{8}{3}
To find B: Substitute (x=-2) (root of (x+2)).
3(-2)+5 = A(-2+2) + B(-2-1)
-6+5 = 0 + B(-3)
-1 = -3B Rightarrow B = frac{1}{3}
The correct coefficients are (A = frac{8}{3}) and (B = frac{1}{3}).
💡 Prevention Tips:
  • Be Meticulous: Perform each arithmetic step slowly and carefully.
  • Double-Check Signs: Pay extra attention to negative numbers, especially when multiplying or subtracting.
  • Verify Solutions: After finding A, B, C, substitute them back into the expanded equation ( 3x+5 = A(x+2) + B(x-1) ) and check if the LHS equals the RHS. This is a quick and effective verification method.
  • Practice Solving Systems: Regularly practice solving simultaneous equations, as this skill is heavily used when equating coefficients.
CBSE_12th
Critical Conceptual

❌ <h3><span style='color: #FF0000;'>Ignoring Improper Rational Functions in Partial Fraction Decomposition</span></h3>

A critical conceptual mistake is directly applying partial fraction decomposition to rational functions where the degree of the numerator is greater than or equal to the degree of the denominator (improper rational functions). Partial fraction decomposition is strictly applicable only to proper rational functions (degree of numerator < degree of denominator). Failing to recognize and address this first step leads to an entirely incorrect decomposition and integral.

💭 Why This Happens:

  • Lack of Fundamental Understanding: Students often overlook the prerequisite condition for partial fraction decomposition.

  • Haste and Pattern Recognition: Rushing to apply learned decomposition patterns without first verifying the nature of the rational function.

  • Algebraic Oversight: Assuming that the standard setup for proper fractions will yield correct coefficients for improper ones, which is algebraically impossible.

✅ Correct Approach:

For both CBSE and JEE, if the rational function is improper, the first and mandatory step is polynomial long division. This converts the improper fraction into a sum of a polynomial and a proper rational function.

The steps are:

  1. Check Degrees: Compare degree(Numerator) and degree(Denominator).

  2. Perform Long Division: If degree(Numerator) ≥ degree(Denominator), divide the numerator by the denominator. Express the function as:
    Rational Function = Quotient + (Remainder / Denominator)

  3. Decompose Proper Fraction: Apply partial fraction methods only to the (Remainder / Denominator) part, which is now a proper rational function.

  4. Integrate All Parts: Integrate the polynomial Quotient term and the decomposed proper fraction terms separately.

📝 Examples:
❌ Wrong:

Consider $int frac{x^2 + 5}{x^2 - 4} dx$.

A common mistake is to directly assume $frac{x^2 + 5}{x^2 - 4} = frac{A}{x-2} + frac{B}{x+2}$. This is fundamentally incorrect because degree(Numerator) = 2 and degree(Denominator) = 2, making it an improper fraction. Setting it up this way will lead to inconsistent equations for A and B or incorrect values.

✅ Correct:

For $int frac{x^2 + 5}{x^2 - 4} dx$:

  1. Identify as Improper: degree(Numerator) = 2, degree(Denominator) = 2.

  2. Perform Long Division:
    $frac{x^2 + 5}{x^2 - 4} = frac{(x^2 - 4) + 9}{x^2 - 4} = 1 + frac{9}{x^2 - 4}$

  3. Decompose Proper Fraction: Now, apply partial fractions to $frac{9}{x^2 - 4}$:
    $frac{9}{(x-2)(x+2)} = frac{A}{x-2} + frac{B}{x+2}$
    Solving for A and B (e.g., by setting x=2 and x=-2): A = 9/4, B = -9/4.

  4. Integrate: The integral becomes:
    $int left(1 + frac{9/4}{x-2} - frac{9/4}{x+2}
    ight) dx$
    $= x + frac{9}{4} ln|x-2| - frac{9}{4} ln|x+2| + C$
    $= x + frac{9}{4} lnleft|frac{x-2}{x+2}
    ight| + C$

💡 Prevention Tips:

  • The Golden Rule: Always, always, *always* check the degrees of the numerator and denominator first. Make this your instinctive first step.

  • Master Long Division: Ensure you are proficient in polynomial long division. This is a foundational skill for this topic.

  • Practice Problems with Diverse Forms: Actively seek out and solve problems involving both proper and improper rational functions to reinforce this concept. Don't shy away from complex looking integrals.

JEE_Main
Critical Calculation

❌ Calculation Errors in Determining Coefficients and Integrating Basic Forms

Students frequently make algebraic calculation errors when solving for the unknown constants (A, B, C, etc.) in the partial fraction decomposition. This often involves sign errors or incorrect arithmetic during substitution or coefficient comparison. A second critical calculation mistake is the incorrect integration of the resulting standard forms, particularly omitting the '1/a' factor when integrating terms like `1/(ax+b)` or misapplying formulas for quadratic factors.
💭 Why This Happens:
These errors stem from a combination of factors: haste during algebraic manipulation, a lack of meticulous double-checking, and sometimes, a superficial understanding of integral formulas (e.g., confusing `∫1/(x+b) dx` with `∫1/(ax+b) dx`). Exam pressure further exacerbates these tendencies, leading to careless mistakes.
✅ Correct Approach:
To avoid these pitfalls, meticulously follow these steps:
  • For Coefficients: Systematically solve for constants using both substitution (for distinct linear factors) and equating coefficients (for repeated/irreducible factors). Perform each algebraic step carefully, paying close attention to signs.
  • For Integration: Always remember the general integral form: ∫ 1/(ax+b) dx = (1/a)ln|ax+b| + C. Similarly, for irreducible quadratic factors like `1/(x^2+a^2)`, the integral is (1/a)arctan(x/a) + C.
📝 Examples:
❌ Wrong:
Consider the integral `∫ (1/(3x+4)) dx`. A common calculation error is to write the integral directly as `ln|3x+4| + C`, completely missing the crucial '1/3' factor. Another example is if `(x+3)/((x-1)(x-2)) = A/(x-1) + B/(x-2)`. If calculating A, substituting `x=1` gives `1+3 = A(1-2)`, which simplifies to `4 = -A`. A common algebraic mistake is to conclude `A=4` instead of `A=-4`.
✅ Correct:
For `∫ (1/(3x+4)) dx`, the correct calculation involves remembering the `1/a` factor, leading to the result: (1/3)ln|3x+4| + C.
For the coefficient calculation: `4 = -A` correctly implies A = -4. Each step must be carefully computed.
💡 Prevention Tips:
  • Systematic Calculation: Always write down algebraic steps clearly when finding coefficients. Do not rush mental calculations.
  • Formula Recall: Memorize and deeply understand the standard integration formulas, especially the `1/a` factor for linear denominators.
  • Double-Check: After finding constants, substitute them back into one or two points to quickly verify. Always review your final integration step for any missing factors or sign errors.
  • Practice: Regular practice with varied problems helps solidify correct calculation habits.
JEE_Main
Critical Other

❌ Skipping the Initial Proper Rational Function Check

A common critical mistake in JEE Advanced is directly applying partial fraction decomposition without first checking if the given rational function is proper (degree of numerator < degree of denominator). If the rational function is improper (degree of numerator ≥ degree of denominator), a mandatory initial step of polynomial long division is often overlooked.
💭 Why This Happens:
This error stems from a lack of complete understanding of the prerequisites for partial fraction decomposition or undue haste. Students often memorize the forms of partial fractions without fully grasping the condition under which they are applicable. They might view it purely as an algebraic manipulation rather than a technique applied to a specific type of rational function.
✅ Correct Approach:
Always compare the degrees of the numerator and denominator. If the degree of the numerator is greater than or equal to the degree of the denominator, perform polynomial long division. The rational function $frac{P(x)}{Q(x)}$ must be rewritten as $S(x) + frac{R(x)}{Q(x)}$, where $S(x)$ is the quotient polynomial and $frac{R(x)}{Q(x)}$ is now a proper rational function (degree of $R(x)$ < degree of $Q(x)$). Partial fractions are then applied only to $frac{R(x)}{Q(x)}$.
📝 Examples:
❌ Wrong:
Consider the integral $int frac{x^3 + 2x^2 - 1}{x^2 - x} dx$. A student might incorrectly attempt to decompose $frac{x^3 + 2x^2 - 1}{x^2 - x}$ directly into $frac{A}{x} + frac{B}{x-1}$. This is fundamentally flawed because the degree of the numerator (3) is greater than the degree of the denominator (2).
✅ Correct:
For the integral $int frac{x^3 + 2x^2 - 1}{x^2 - x} dx$:

  1. Perform polynomial long division:
    $frac{x^3 + 2x^2 - 1}{x^2 - x} = (x + 3) + frac{3x - 1}{x^2 - x}$

  2. Rewrite the integral:
    $int left(x + 3 + frac{3x - 1}{x^2 - x}
    ight) dx$

  3. Apply partial fractions only to the proper rational part:
    $frac{3x - 1}{x(x-1)} = frac{A}{x} + frac{B}{x-1}$. Solving gives $A=1, B=2$.

  4. Integrate the components:
    $int (x+3) dx + int frac{1}{x} dx + int frac{2}{x-1} dx = frac{x^2}{2} + 3x + ln|x| + 2ln|x-1| + C$.

💡 Prevention Tips:

  • Golden Rule: Always check the degrees of the numerator and denominator before starting partial fraction decomposition.

  • If degree(Numerator) $ge$ degree(Denominator), perform polynomial long division first.

  • Remember the structure: $frac{ ext{Improper Rational Function}}{ ext{Integral}} = int ext{(Quotient Polynomial)} dx + int ext{(Proper Rational Remainder)} dx$.

  • JEE Advanced Specific: This initial step is a common trap to test conceptual clarity and can significantly alter the complexity and correctness of the final answer. Master polynomial long division for this purpose.

JEE_Advanced
Critical Approximation

❌ Incorrect Partial Fraction Decomposition Form for Repeated or Irreducible Factors

Students often make critical errors in setting up the correct form of partial fraction decomposition, particularly when the denominator contains repeated linear factors or irreducible quadratic factors. This isn't an approximation in the numerical sense, but an incorrect algebraic 'approximation' of the decomposition structure, leading to an unsolvable or incorrectly solved integral. This error is of critical severity in JEE Advanced.
💭 Why This Happens:
This mistake stems from a superficial understanding of the rules for different types of denominator factors. Haste, misidentification of factor types (e.g., treating `(x-a)^2` as two distinct `(x-a)` factors instead of a repeated one), or failing to verify if a quadratic factor is irreducible often lead to this error. Lack of rigorous practice in setting up diverse partial fraction forms exacerbates the issue.
✅ Correct Approach:
Always identify the type of factor in the denominator first. The correct forms are as follows:
  • Distinct Linear Factor (ax+b): A/(ax+b)
  • Repeated Linear Factor (ax+b)n: A1/(ax+b) + A2/(ax+b)2 + ... + An/(ax+b)n
  • Irreducible Quadratic Factor (ax2+bx+c): (Ax+B)/(ax2+bx+c) (where b2-4ac < 0)
  • Repeated Irreducible Quadratic Factor (ax2+bx+c)n: (A1x+B1)/(ax2+bx+c) + ... + (Anx+Bn)/(ax2+bx+c)n
📝 Examples:
❌ Wrong:
Consider the integral ∫(x+1)/((x-1)2(x+2)) dx. A common wrong decomposition attempt is:
(x+1)/((x-1)2(x+2)) = A/(x-1) + B/(x+2)
This setup completely ignores the repeated nature of `(x-1)2`, making it impossible to correctly determine the coefficients and leading to an incorrect integration.
✅ Correct:
For the same integral ∫(x+1)/((x-1)2(x+2)) dx, the correct partial fraction decomposition form is:
(x+1)/((x-1)2(x+2)) = A/(x-1) + B/((x-1)2) + C/(x+2)
Each term in the denominator's power gets its own fraction, ensuring all coefficients can be uniquely determined and the integral can be correctly evaluated.
💡 Prevention Tips:
  • Rigorous Factorization: Always completely factorize the denominator and identify the type of each factor (linear, repeated linear, irreducible quadratic).
  • Memorize Forms: Understand and commit to memory the specific partial fraction forms for each type of factor.
  • Double-Check Setup: Before proceeding with finding coefficients, re-verify your decomposition setup against the rules.
  • Practice Varied Problems: Work through examples involving all types of factors, especially combinations, to build intuition and accuracy.
JEE_Advanced
Critical Sign Error

❌ Sign Errors in Partial Fraction Coefficient Determination

Students frequently mismanage negative signs when determining coefficients (A, B, C) or during the recombination phase of partial fraction decomposition. This leads to incorrect signs in integrated terms, critically affecting the final answer for JEE Advanced problems.
💭 Why This Happens:
  • Rushed Algebra: Overlooking negative signs during expansion or solving for coefficients.
  • Distribution Errors: Incorrectly applying negative signs, e.g., with `B(x-a)` when 'B' itself is negative or 'a' is negative.
✅ Correct Approach:
Employ a meticulous, step-by-step method to prevent sign errors:
  1. Systematic Algebra: Double-check every expansion, combination, and substitution step for signs when determining coefficients.
  2. Precise Integration: Ensure each partial fraction term is integrated with its exact determined sign. For example, if a coefficient `B` is `-3/4`, the term should correctly be `- (3/4)ln|x+2|`.
📝 Examples:
❌ Wrong:
Consider the integral: ∫ (x+5) / (x2 - 4) dx.
After setting (x+5) / ((x-2)(x+2)) = A/(x-2) + B/(x+2), solving x+5 = A(x+2) + B(x-2) yields A = 7/4 and B = -3/4. A common sign error is to incorrectly write the decomposed form by losing the negative sign for B, e.g., (7/4)/(x-2) + (3/4)/(x+2). This leads to the wrong integral: (7/4)ln|x-2| + (3/4)ln|x+2| + C.
✅ Correct:
For the same integral, using the correct coefficients A = 7/4 and B = -3/4, the correct partial fraction decomposition is: (7/4)/(x-2) + (-3/4)/(x+2). The correct integration then becomes:
∫ [ (7/4)/(x-2) - (3/4)/(x+2) ] dx = (7/4)ln|x-2| - (3/4)ln|x+2| + C.
Notice the crucial negative sign for the second logarithmic term, which is vital for the correct solution.
💡 Prevention Tips:
  • Clear Layout: Use sufficient space for calculations; avoid crowding steps.
  • Verify Coefficients: Double-check the sign of each A, B, C value immediately after calculation.
  • Meticulous Substitution: Ensure all terms and their signs are handled precisely when substituting 'x' values.
  • Final Review: Before integrating, scrupulously check the entire decomposed expression for any sign errors.
JEE_Advanced
Critical Unit Conversion

❌ Incorrect Partial Fraction Form Setup

Students frequently make critical errors in setting up the correct algebraic form of partial fractions based on the nature of the denominator's factors. This is analogous to a fundamental 'conversion' error, as an incorrect setup renders all subsequent calculations invalid. Common issues include misidentifying repeated factors, overlooking irreducible quadratic factors, or failing to handle improper rational functions first.
💭 Why This Happens:
This mistake stems from a lack of complete understanding of the various cases of partial fraction decomposition. Students often rush the setup, confuse distinct linear factors with repeated ones, or incorrectly decompose irreducible quadratic factors. Sometimes, they forget to perform polynomial long division when the degree of the numerator is greater than or equal to the degree of the denominator (improper rational function).
✅ Correct Approach:
Always begin by ensuring the rational function is proper (degree of numerator < degree of denominator). If not, perform polynomial long division first. Then, factor the denominator completely into linear and/or irreducible quadratic factors.
Apply the correct partial fraction form for each factor type:

  • Distinct Linear Factor (ax+b): A/(ax+b)
  • Repeated Linear Factor (ax+b)^n: A1/(ax+b) + A2/(ax+b)^2 + ... + An/(ax+b)^n
  • Distinct Irreducible Quadratic Factor (ax^2+bx+c): (Ax+B)/(ax^2+bx+c)
  • Repeated Irreducible Quadratic Factor (ax^2+bx+c)^n: (A1x+B1)/(ax^2+bx+c) + (A2x+B2)/(ax^2+bx+c)^2 + ... + (Anx+Bn)/(ax^2+bx+c)^n
📝 Examples:
❌ Wrong:

Consider the integral ∫ (x+1) / (x(x-1)2) dx

Incorrect Setup: (x+1) / (x(x-1)2) = A/x + B/(x-1)
Here, the repeated linear factor (x-1)2 is incorrectly treated as a distinct linear factor (x-1), omitting the term B/(x-1)2.

✅ Correct:

For the integral ∫ (x+1) / (x(x-1)2) dx

Correct Setup: (x+1) / (x(x-1)2) = A/x + B/(x-1) + C/(x-1)2
This correctly accounts for the distinct linear factor 'x' and the repeated linear factor '(x-1)2'.

💡 Prevention Tips:
  • Memorize Forms: Understand and commit to memory the distinct partial fraction forms for each factor type.
  • Factor Carefully: Always factor the denominator completely and accurately.
  • Check Degree: Before setting up, compare the degrees of the numerator and denominator. Perform polynomial long division if the function is improper.
  • Practice Variety: Solve problems involving all types of factors (distinct linear, repeated linear, irreducible quadratic, repeated irreducible quadratic) to solidify understanding.
  • JEE Advanced Tip: Be especially vigilant with complex denominators involving combinations of factor types; these are often tested.
JEE_Advanced
Critical Formula

❌ Incorrect Partial Fraction Form for Irreducible Quadratic Factors

Students often misapply the partial fraction decomposition rules when the denominator contains an irreducible quadratic factor (a quadratic that cannot be factored into real linear factors). The common error is to assign a constant (e.g., A) as the numerator instead of a linear term (e.g., Ax + B).
💭 Why This Happens:
This mistake stems from a lack of complete understanding of the fundamental principle behind partial fraction decomposition: the degree of the numerator for each partial fraction term must be one less than the degree of its corresponding denominator factor. For an irreducible quadratic factor (degree 2), the numerator must therefore be a linear polynomial (degree 1). Students might confuse this rule with linear factors which correctly take a constant numerator.
✅ Correct Approach:
For every irreducible quadratic factor of the form (ax² + bx + c) in the denominator, the corresponding partial fraction must have a linear polynomial in its numerator. This form is (Ax + B) / (ax² + bx + c), where A and B are constants to be determined.
📝 Examples:
❌ Wrong:
For the integral ∫ (x / ((x-1)(x² + 4))) dx , a common incorrect decomposition is:
x / ((x-1)(x² + 4)) = A / (x-1) + B / (x² + 4)
This assigns a constant 'B' to the irreducible quadratic factor (x² + 4), which is wrong.
✅ Correct:
The correct decomposition for the integral ∫ (x / ((x-1)(x² + 4))) dx is:
x / ((x-1)(x² + 4)) = A / (x-1) + (Bx + C) / (x² + 4)
Here, (Bx + C) is the correct linear numerator for the irreducible quadratic factor (x² + 4).
💡 Prevention Tips:
  • Understand the Rule: Always ensure the numerator's degree is one less than the denominator's degree for each partial fraction term.
  • Identify Irreducible Quadratics: Before decomposing, check if quadratic factors like (x² + a²) or (ax² + bx + c) (where discriminant D < 0) are truly irreducible.
  • Practice Diverse Problems: Work through problems involving all types of factors (linear, repeated linear, irreducible quadratic) to solidify the correct forms.
  • JEE Advanced Note: This error is critical as it will lead to an incorrect system of equations for coefficients and ultimately an unintegrable form or wrong answer.
JEE_Advanced
Critical Calculation

❌ Incorrect Calculation of Coefficients in Partial Fraction Decomposition

A frequent critical error in integration by partial fractions is making algebraic mistakes while determining the unknown coefficients (A, B, C, etc.). Whether using the equating coefficients method or the substitution method (like Heaviside's cover-up), calculation inaccuracies in finding these values lead to an incorrect partial fraction setup and, consequently, an erroneous final integral.
💭 Why This Happens:
This mistake primarily stems from:
  • Algebraic carelessness: Errors in handling negative signs, fractions, or solving simultaneous equations.
  • Rushing the process: Especially under exam pressure, students might overlook small calculation errors.
  • Incomplete verification: Not cross-checking the calculated coefficients, which is a crucial step for JEE Advanced.
✅ Correct Approach:
To accurately find the coefficients:
  • Systematic Setup: Ensure the partial fraction form is correctly established for the given denominator (e.g., linear distinct, repeated, irreducible quadratic factors).
  • Equating Coefficients: Multiply both sides by the original denominator. Expand the right side and equate coefficients of like powers of 'x' (e.g., x², x, constant term) on both sides. Solve the resulting system of linear equations carefully.
  • Substitution Method (Heaviside's Cover-up): For distinct linear factors, substitute the roots of the denominator into the expanded equation to directly solve for the respective coefficients. This is generally faster for linear factors.
  • Verification (Crucial for JEE Advanced): After calculating all coefficients, pick a simple, non-root value for 'x' and substitute it into both the original rational function and your partial fraction decomposition. The results must match. This immediately highlights any calculation errors.
📝 Examples:
❌ Wrong:
Consider ∫ (5x - 7) / (x - 1)(x + 2) dx.
Decomposition: (5x - 7) / (x - 1)(x + 2) = A / (x - 1) + B / (x + 2)
Expanding: 5x - 7 = A(x + 2) + B(x - 1)
If a student correctly finds A = -2/3 by setting x = 1 (5(1) - 7 = A(1+2) → -2 = 3A → A = -2/3), but then makes a sign error for B:
Let x = -2: 5(-2) - 7 = B(-2 - 1) → -10 - 7 = -3B → -17 = -3B → B = -17/3 (Incorrectly writing B = 17/3 due to sign mistake).
This single calculation error leads to an incorrect integral.
✅ Correct:
Continuing the above example for ∫ (5x - 7) / (x - 1)(x + 2) dx:
Decomposition: (5x - 7) / (x - 1)(x + 2) = A / (x - 1) + B / (x + 2)
Expanding: 5x - 7 = A(x + 2) + B(x - 1)
  • To find A: Set x = 1.
    5(1) - 7 = A(1 + 2) + B(1 - 1) ⇒ -2 = 3A ⇒ A = -2/3
  • To find B: Set x = -2.
    5(-2) - 7 = A(-2 + 2) + B(-2 - 1) ⇒ -17 = -3B ⇒ B = 17/3
Verification: Let x = 0 (a non-root value).
Original: (5(0) - 7) / ((0 - 1)(0 + 2)) = -7 / (-2) = 7/2
Partial Fractions: (-2/3) / (0 - 1) + (17/3) / (0 + 2) = 2/3 + 17/6 = 4/6 + 17/6 = 21/6 = 7/2
The values match, confirming the coefficients are correct.
Then, ∫ [ (-2/3) / (x - 1) + (17/3) / (x + 2) ] dx = (-2/3) ln|x - 1| + (17/3) ln|x + 2| + C
💡 Prevention Tips:
  • Be Meticulous with Algebra: Pay close attention to signs, fractions, and solving simultaneous equations. Write down each step clearly.
  • Systematic Approach: Follow a consistent method for finding coefficients (e.g., always use substitution for linear factors, then equating coefficients for remaining terms).
  • Employ Verification (JEE Advanced specific): Always cross-check your calculated coefficients by substituting a convenient non-root value of 'x' into both the original function and the partial fraction form. This is the most effective way to catch calculation errors before proceeding to integration.
  • Practice Diverse Problems: Work through examples involving distinct linear, repeated linear, and irreducible quadratic factors to gain proficiency in coefficient determination.
JEE_Advanced
Critical Conceptual

❌ Ignoring Proper Fraction Condition and Incorrect Partial Fraction Form Setup

Students frequently overlook the fundamental condition that partial fraction decomposition is directly applicable only to proper rational functions (degree of numerator < degree of denominator). If the function is improper, polynomial division must be performed first. Additionally, they often misidentify or incorrectly apply the decomposition form for different types of denominator factors (e.g., irreducible quadratic factors, repeated linear factors).
💭 Why This Happens:
This mistake stems from a rush to apply the partial fraction technique without checking its prerequisites. A conceptual gap in understanding the algebraic structure required for decomposition, especially for complex denominators, and insufficient practice in setting up the correct forms are common reasons. Students might also confuse the forms for distinct linear, repeated linear, and irreducible quadratic factors.
✅ Correct Approach:
Always begin by comparing the degrees of the numerator and denominator. If the rational function is improper (), perform polynomial long division to express it as a polynomial plus a proper rational function: $$frac{P(x)}{Q(x)} = S(x) + frac{R(x)}{Q(x)}$$ Then, apply partial fraction decomposition only to the proper fraction $frac{R(x)}{Q(x)}$. For the decomposition itself, ensure the correct form is used for each factor type in the denominator:
📝 Examples:
❌ Wrong:
For $int frac{x^3+x^2+1}{x^2+1} dx$, incorrectly attempting to decompose directly as $int left(frac{Ax+B}{x^2+1}
ight) dx$. This is wrong because the integrand is an improper rational function (degree of numerator = 3, degree of denominator = 2).
✅ Correct:
For $int frac{x^3+x^2+1}{x^2+1} dx$:
1. Perform polynomial division: $$(x^3+x^2+1) div (x^2+1) = x+1 + frac{-x}{x^2+1}$$
2. The integral becomes: $$int left(x+1 - frac{x}{x^2+1}
ight) dx$$
3. Integrate each term: $$frac{x^2}{2} + x - frac{1}{2} ln|x^2+1| + C$$
This correctly handles the improper fraction. JEE Advanced Note: For denominators like $(x^2+x+1)(x-1)^2$, the correct form would be $frac{Ax+B}{x^2+x+1} + frac{C}{x-1} + frac{D}{(x-1)^2}$. Missing the $(x-1)$ term or using $A/(x^2+x+1)$ are common errors.
💡 Prevention Tips:
  • Always check degrees first: Make it a habit to compare the degrees of the numerator and denominator before starting any partial fraction decomposition.
  • Understand the four standard forms of partial fractions thoroughly (linear, repeated linear, irreducible quadratic, repeated irreducible quadratic).
  • Practice setting up the decomposition for various types of denominator factors.
  • For JEE Advanced: Pay close attention to composite denominators involving multiple factor types, as incorrect setup is a frequent source of error.
JEE_Advanced
Critical Formula

❌ Incorrect Partial Fraction Decomposition Setup

Students frequently err in setting up the partial fraction decomposition for rational functions, especially when the denominator contains repeated linear factors or irreducible quadratic factors. Instead of assigning a unique term for each power of a repeated factor or using an (Ax+B) form for irreducible quadratics, they might simplify the setup incorrectly, leading to an insoluble system of equations for coefficients or an incorrect final integral.
💭 Why This Happens:
This critical mistake stems from a fundamental misunderstanding of the rules governing partial fraction decomposition. Students often treat repeated factors as distinct linear factors or incorrectly assume a constant numerator for irreducible quadratic factors, failing to recall the specific forms required for each type of factor. Insufficient practice in recognizing and applying these specific forms contributes significantly.
✅ Correct Approach:
Always analyze the denominator's factors carefully and apply the correct decomposition rule:
  • For a non-repeated linear factor (ax+b): A/(ax+b)
  • For a repeated linear factor (ax+b)n: A1/(ax+b) + A2/(ax+b)2 + ... + An/(ax+b)n
  • For a non-repeated irreducible quadratic factor (ax2+bx+c): (Ax+B)/(ax2+bx+c)
JEE Specific: Mastering these setups is crucial as complex denominators are common, and an incorrect initial setup makes the entire problem unsolvable.
📝 Examples:
❌ Wrong:
Consider the integral ∫ (x+5) / (x(x-2)2(x2+1)) dx.
A common wrong setup for partial fractions would be:
A/x + B/(x-2) + C/(x2+1)
This setup ignores the repeated nature of (x-2) and uses an incorrect numerator for the irreducible quadratic factor (x2+1).
✅ Correct:
For the same integral ∫ (x+5) / (x(x-2)2(x2+1)) dx,
The correct partial fraction decomposition is:
A/x + B/(x-2) + C/(x-2)2 + (Dx+E)/(x2+1)
This correct setup ensures all coefficients can be determined and the integral can be solved.
💡 Prevention Tips:
  • Memorize Forms: Thoroughly learn and understand the four fundamental forms of partial fraction decomposition.
  • Factorize First: Always ensure the denominator is completely factorized before attempting decomposition.
  • Identify Factor Types: Clearly identify if factors are linear, repeated linear, irreducible quadratic, or repeated irreducible quadratic.
  • Practice Setup: Practice setting up the decomposition for various complex denominators, even if you don't solve for coefficients. This builds muscle memory for the correct structure.
JEE_Main
Critical Unit Conversion

❌ Ignoring Proper/Improper Rational Functions & Incorrect Decomposition Setup

A common and critical mistake students make in Integration by Partial Fractions is failing to first identify if the given rational function is proper or improper. If the degree of the numerator is greater than or equal to the degree of the denominator, it's an improper fraction and requires polynomial long division before applying partial fractions. Another critical error is the incorrect setup of the partial fraction decomposition, especially for repeated linear factors or irreducible quadratic factors in the denominator. This fundamental algebraic error renders all subsequent integration steps incorrect.

Note on 'Unit Conversion': It's important to clarify that 'unit conversion' is a concept related to physical quantities and their dimensions, which is not applicable to the mathematical technique of integration by partial fractions. This section focuses on conceptual misunderstandings within the algebraic decomposition process itself.
💭 Why This Happens:
  • Lack of clear understanding of the preconditions for applying partial fractions (i.e., the function must be proper).
  • Confusion or misremembering the rules for setting up partial fractions for different types of factors (e.g., simple linear, repeated linear, irreducible quadratic).
  • Haste in problem-solving, leading to skipping the initial check or misidentifying factor types.
  • Weak algebraic manipulation skills in setting up and solving for constants.
✅ Correct Approach:
  1. Check Degree: Always compare the degree of the numerator (N(x)) with the degree of the denominator (D(x)).
    • If deg(N(x)) ≥ deg(D(x)) (improper fraction), perform polynomial long division first: N(x)/D(x) = Q(x) + R(x)/D(x), where deg(R(x)) < deg(D(x)). Then apply partial fractions to R(x)/D(x).
    • If deg(N(x)) < deg(D(x)) (proper fraction), proceed directly to decomposition.
  2. Factor Denominator: Fully factorize the denominator into linear and/or irreducible quadratic factors.
  3. Setup Decomposition: Apply the correct partial fraction form based on the factors:
    • Linear factor (ax+b): A/(ax+b)
    • Repeated linear factor (ax+b)^n: A1/(ax+b) + A2/(ax+b)^2 + ... + An/(ax+b)^n
    • Irreducible quadratic factor (ax^2+bx+c): (Ax+B)/(ax^2+bx+c)
    • Repeated irreducible quadratic factor (ax^2+bx+c)^n: (A1x+B1)/(ax^2+bx+c) + ... + (Anx+Bn)/(ax^2+bx+c)^n
  4. Solve for Constants: Use methods like equating coefficients or substitution of roots to find the values of A, B, C, etc.
📝 Examples:
❌ Wrong:

Consider the integral: ∫ [x^2 / ((x-1)^2 * (x+2))] dx

Wrong Setup: A student might incorrectly decompose x^2 / ((x-1)^2 * (x+2)) as:

A/(x-1) + B/(x+2)

This is wrong because it misses the term for the repeated linear factor and incorrectly handles the degree of the numerator which is equal to the degree of the denominator (improper fraction if considering only the fractional part after long division).

✅ Correct:

Consider the integral: ∫ [x^2 / ((x-1)^2 * (x+2))] dx

Step 1: Check Degree (CBSE & JEE): Degree of Numerator (x^2) = 2. Degree of Denominator ((x-1)^2 * (x+2) = x^3 - 3x + 2) = 3. Since 2 < 3, it's a proper fraction. No long division needed here.

Step 2: Correct Decomposition Setup (CBSE & JEE): For the denominator (x-1)^2 * (x+2) (repeated linear factor (x-1) and simple linear factor (x+2)), the correct partial fraction decomposition is:

x^2 / ((x-1)^2 * (x+2)) = A/(x-1) + B/(x-1)^2 + C/(x+2)

Then, multiply by the denominator to solve for A, B, C:

x^2 = A(x-1)(x+2) + B(x+2) + C(x-1)^2

By substituting values (e.g., x=1, x=-2) and equating coefficients, one can find A, B, C.

💡 Prevention Tips:
  • Master Preconditions: Always start by checking if the rational function is proper. If not, perform polynomial long division. This is non-negotiable for both CBSE and JEE.
  • Learn Decomposition Rules by Heart: Create a cheat sheet or flashcards for the four types of factor decompositions.
  • Practice Setup Extensively: Focus on problems that test different combinations of factors. The setup is half the battle won.
  • Double-Check Your Work: Before solving for constants, quickly re-verify if your partial fraction form matches the denominator's factors.
  • JEE Specific: In JEE, often the problems might involve complex factorizations, requiring strong algebraic skills to even get to the partial fraction stage. Don't rush factorization.
JEE_Main
Critical Other

❌ Incorrect Setup of Partial Fractions for Repeated or Irreducible Quadratic Factors

A common critical error is failing to correctly set up the partial fraction decomposition for rational functions where the denominator contains repeated linear factors (e.g., (ax+b)n) or irreducible quadratic factors (e.g., (ax2+bx+c)). This algebraic error propagates throughout the integration process, leading to a completely wrong final answer.
💭 Why This Happens:
This mistake stems from a superficial understanding of partial fraction rules beyond the simplest case of distinct linear factors. Students often rush the decomposition step under exam pressure, or confuse the specific forms required for repeated factors with those for distinct ones. Forgetting to provide a linear numerator (Ax+B) for irreducible quadratic factors is also a frequent oversight.
✅ Correct Approach:
The setup for partial fraction decomposition must strictly follow these rules:
  • For each repeated linear factor (ax+b)n in the denominator, include n terms: A1/(ax+b) + A2/(ax+b)2 + ... + An/(ax+b)n.
  • For each irreducible quadratic factor (ax2+bx+c) (where b2-4ac < 0), the corresponding term in the decomposition must be (Ax+B)/(ax2+bx+c).
  • If an irreducible quadratic factor is repeated, e.g., (ax2+bx+c)n, the decomposition includes terms like (A1x+B1)/(ax2+bx+c) + (A2x+B2)/(ax2+bx+c)2 + ... + (Anx+Bn)/(ax2+bx+c)n.
📝 Examples:
❌ Wrong:
When integrating ∫ (x+1) / (x2(x2+1)) dx, an incorrect decomposition would be to write it as A/x + B/(x2+1). This misses the term for x2 (the repeated linear factor (x)2) and provides an incorrect numerator form for the irreducible quadratic factor (x2+1).
✅ Correct:
For the integral ∫ (x+1) / (x2(x2+1)) dx, the correct partial fraction decomposition is:
(x+1) / (x2(x2+1)) = A/x + B/x2 + (Cx+D)/(x2+1)
This correctly accounts for the repeated linear factor (x)2 and the irreducible quadratic factor (x2+1) with its linear numerator.
💡 Prevention Tips:
  • Master the Rules: Thoroughly understand and memorize the specific setup rules for all types of factors (distinct linear, repeated linear, irreducible quadratic, repeated irreducible quadratic).
  • Systematic Approach: Before decomposing, always ensure the degree of the numerator is less than the denominator's degree (perform long division if not). Then, factorize the denominator completely.
  • Practice Complex Cases: JEE Main frequently tests these more complex partial fraction setups. Practice problems involving repeated factors and irreducible quadratic factors to solidify your understanding. CBSE usually focuses on simpler cases, so extra JEE-specific practice is vital.
  • Double-Check: After setting up the decomposition, quickly review it against the rules to catch any obvious errors before proceeding to solve for the constants.
JEE_Main

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Integration by partial fractions

Subject: Mathematics
Unit: 8 - INTEGRAL CALCULUS
Sub-unit: 8.2 - Techniques of Integration
Complexity: High
Syllabus: JEE_Main

Content Completeness: 55.6%

55.6%
📚 Explanations: 0
📝 CBSE Problems: 18
🎯 JEE Problems: 19
🎥 Videos: 0
🖼️ Images: 0
📐 Formulas: 5
📚 References: 10
⚠️ Mistakes: 59
🤖 AI Explanation: No