Integration by Partial Fractions - Definition, Formulas, Steps and Examples

Meaning of Integration and Rational Functions

Integration is the reverse of differentiation. When you integrate a function, you’re finding the original expression whose rate of change is given. If $y$ changes with respect to $x$, then its derivative represents the slope of the tangent at any point.

A rational function is a function of the form
$ \frac{P(x)}{Q(x)} $,
where $P(x)$ and $Q(x)$ are polynomials and $Q(x) \neq 0$.

Partial fractions are used only when a rational function can be broken into simpler fractions.

The expression $ \frac{P(x)}{Q(x)} $ is decomposed into simpler rational expressions whose integrals are easy to compute. This breakdown is called partial fraction decomposition.

This method applies only when $ \deg(P(x)) < \deg(Q(x)) $. If the function is improper (degree of numerator ≥ degree of denominator), we begin by performing long division.

When Degree of Numerator ≥ Degree of Denominator

To integrate expressions where $ \deg(P(x)) \geq \deg(Q(x)) $, divide first.

Example:
Evaluate
$ \int \frac{x^2 + 3x + 5}{x + 1} , dx $

Since $ \deg(x^2 + 3x + 5) \geq \deg(x+1) $, perform long division:

$ \frac{x^2 + 3x + 5}{x+1} = x + 2 + \frac{3}{x+1} $

Now integrate:

$ \int \left( x + 2 + \frac{3}{x+1} \right) dx
= \frac{1}{2}x^2 + 2x + 3 \ln|x+1| + C $

Case 1: Denominator with Non-Repeated Linear Factors

If the denominator splits like
$ Q(x) = (x - a_1)(x - a_2)(x - a_3)\cdots(x - a_n) $,
then the decomposition is:

$ \frac{P(x)}{Q(x)} =
\frac{A_1}{x - a_1} + \frac{A_2}{x - a_2} + \cdots + \frac{A_n}{x - a_n} $

The constants $A_1, A_2, \ldots, A_n$ are determined by:

• equating numerators, or

• substituting $x = a_1, a_2, \ldots, a_n$ (shortcut method)

Shortcut Example

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

Put $x=0,1,2$ to get $A,B,C$ quickly.

Illustration 1

Evaluate
$ \int \frac{3x + 2}{x^3 - x^2 - 2x} , dx $

Factor the denominator:

$ x^3 - x^2 - 2x = x(x-2)(x+1) $

So,
$ \frac{3x+2}{x(x-2)(x+1)}
= \frac{A}{x} + \frac{B}{x-2} + \frac{C}{x+1} $

Cross-multiplying:

$ 3x + 2 = A(x-2)(x+1) + Bx(x+1) + Cx(x-2) $

Now substitute:

• $x = 0 \Rightarrow A = -1$

• $x = 2 \Rightarrow B = \frac{4}{3}$

• $x = -1 \Rightarrow C = -\frac{1}{3}$

Rewrite the integral:

$ \int \left( -\frac{1}{x} + \frac{4}{3} \cdot \frac{1}{x-2} - \frac{1}{3} \cdot \frac{1}{x+1} \right) dx $

Integrating:

$ -\ln|x| + \frac{4}{3} \ln|x-2| - \frac{1}{3}\ln|x+1| + C $

Case 2: Denominator with Repeated Linear Factors

If
$ Q(x) = (x-a)^k(x-a_1)(x-a_2)\cdots(x-a_n) $,
the decomposition becomes:

$
\frac{P(x)}{Q(x)} =
\frac{A_1}{x-a} + \frac{A_2}{(x-a)^2} + \cdots + \frac{A_k}{(x-a)^k}

• \frac{B_1}{x - a_1} + \frac{B_2}{x - a_2} + \cdots + \frac{B_n}{x - a_n}
  $

Each term is integrated separately.

Case 3: Denominator with Irreducible Quadratic Factors

For each quadratic factor
$ ax^2 + bx + c $
(of the type that cannot be factored further), assume:

$ \frac{Ax + B}{ax^2 + bx + c} $

These integrate using substitution or logarithm/arctan rules.

Steps to Resolve an Expression into Partial Fractions

Integration by partial fractions is used when the integrand is a rational function, i.e., a ratio of two polynomials.

Step 1: Ensure the Rational Function Is Proper

Check the degree of numerator and denominator.

• If degree of numerator < degree of denominator → proceed

• If degree of numerator ≥ degree of denominator → first perform long division

Step 2: Factor the Denominator Completely

Factor the denominator into linear and/or quadratic factors.

Examples:

• $(x+1)(x+2)$

• $(x-1)^2$

• $(x^2+1)$

Step 3: Write Partial Fractions Based on Factor Type

Use standard forms:

For distinct linear factors:
$\dfrac{A}{x-a} + \dfrac{B}{x-b}$

For repeated linear factors:
$\dfrac{A}{x-a} + \dfrac{B}{(x-a)^2}$

For irreducible quadratic factors:
$\dfrac{Ax+B}{x^2+ax+b}$

Step 4: Find the Constants

Multiply both sides by the denominator and compare coefficients or substitute suitable values of $x$ to find constants like $A$, $B$, etc.

Integration by Partial Fractions – Formula and Method

Once the expression is resolved into partial fractions, integrate each term separately using standard integration formulas.

If
$\dfrac{P(x)}{Q(x)} = \dfrac{A}{x-a} + \dfrac{B}{x-b}$

then
$\int \dfrac{P(x)}{Q(x)} dx = \int \dfrac{A}{x-a} dx + \int \dfrac{B}{x-b} dx$

This method simplifies complex rational expressions and is widely used in Class 12 board exams and competitive exams.

Solving Integrals Using Partial Fractions (Step-by-Step)

We have given below the step by step method to solve the integrals by using partial fractions:

Step 1: Express the Integral in Rational Form

Write the given integral as
$\int \dfrac{P(x)}{Q(x)} dx$

Step 2: Resolve into Partial Fractions

Decompose the rational function using the steps discussed earlier.

Step 3: Integrate Each Term Separately

Apply basic integration formulas such as:

$\int \dfrac{1}{x-a} dx = \ln|x-a|$

$\int \dfrac{1}{x^2+a^2} dx = \dfrac{1}{a}\tan^{-1}\left(\dfrac{x}{a}\right)$

Step 4: Add Constant of Integration

For indefinite integrals, always write $+ C$

Solved Examples Based On Integration using Partial Function:

Example 1: Integrate $\int \frac{x^3-x}{x^2+1} d x$

1) ${x+\ln \left(x^2+1\right)+C}$
2) $\frac{x^2}{2}+\ln \left(x^2+1\right)+C$
3) $x-\ln \left(x^2+1\right)+C$
4) $\frac{x^2}{2}-\ln \left(x^2+1\right)+C$

Solution

$\begin{aligned}
& \int \frac{x\left(x^2-1\right) d x}{x^2+1} \\
& \begin{aligned}
\Rightarrow \int \frac{x\left(x^2+1\right)-2 x}{x^2+1} d x & =\int x d x-\int \frac{2 x}{x^2+1} d x \\
& =\frac{x^2}{2}-\ln \left(x^2+1\right)+C
\end{aligned}
\end{aligned}$
Hence, the answer is the option (4).

Example 2:Integrate $\int \frac{x^3+x^3+x}{x^2+1} d x$
1) $\frac{x^3}{3}+\ln \left(x^2+1\right)+C$
2) $\frac{x^4}{3}+\ln \left(x^2+1\right)+C$
3) $\frac{x^4}{4}+\frac{1}{2} \ln \left(x^2+1\right)+C$
4) none of these

Solution

$\begin{aligned}
\int \frac{x^5+x^3+x}{x^2+1} d x= & \int \frac{x^3\left(x^2+1\right)}{x^2+1} d x+\int \frac{x}{x^2+1} d x \\
& =\frac{x^4}{4}+\frac{1}{2} \ln \left(x^2+1\right)+C
\end{aligned}$

Hence, the answer is the option (3).

Example 3: If $f\left(\frac{3 x-4}{3 x+4}\right)=x+2, x \neq-\frac{4}{3}$ and $\int f(x) d x=A \log |1-x|+B x+C$ then the ordered pair (A, B ) is equal to: (where C is a constant of integration)
1) $\left(\frac{8}{3}, \frac{2}{3}\right)$
2) $\left(-\frac{8}{3}, \frac{2}{3}\right)$
3) $\left(-\frac{8}{3},-\frac{2}{3}\right)$
4) $\left(\frac{8}{3},-\frac{2}{3}\right)$

Solution

As learned in the concept

Rule of integration by Partial fraction -

Linear and non-repeated:

$\begin{aligned}
& \frac{P(x)}{Q(x)}=\frac{P(x)}{\left(x-\alpha_1\right)\left(x-\alpha_2\right) \cdots\left(x-\alpha_n\right)} \\
& \text { Let } \frac{P(x)}{Q(x)}=\frac{A}{\left(x-\alpha_1\right)}+\frac{B}{\left(x-\alpha_2\right)} \cdots
\end{aligned}$
Find $A, B \ldots$
By comparing $N^r$ and $P(x)$

$f\left(\frac{3 x-4}{3 x+4}\right)=x+2$
Put $\frac{3 x-4}{3 x+4}=y$

$\begin{aligned}
& =>3 x y+4 y=3 x-4 \\
& x=\frac{-4(y+1)}{3(y-1)} \\
& x=\frac{4(1+y)}{3(1-y)} \\
& f(y)=\frac{4(1+y)}{3(1-y)}+2 \\
& =\frac{10-2 y}{3(1-y)}
\end{aligned}$
$\begin{aligned}
& \therefore f(x)=\frac{2(5-x)}{3(1-x)} \\
& \therefore \frac{2}{3} \int\left(\frac{5-x}{1-x}\right) d x=\frac{2}{3} \int \frac{1-x+4}{(1-x)} d x \\
& =\frac{2}{3} x+\frac{8}{3} \ln |1-x|+C
\end{aligned}$
So, $A=8 / 3$ and $B=2 / 3$
Hence, the answer is the option 1.
Example 4: Find $\int \frac{6}{(x-1)(x+1)} d x$ :
1) $3 \ln (x+1)(x-1)+C$

2) $3 \ln (x+1) /(x-1)+C$

3) $3 \ln (x-1)-3 \ln (x+1)+C$

4) none of these

Solution

As we have learned

Rule of integration by Partial fraction -

Linear and non-repeated:
$
\begin{aligned}
& \frac{P(x)}{Q(x)}=\frac{P(x)}{\left(x-\alpha_1\right)\left(x-\alpha_2\right) \cdots\left(x-\alpha_n\right)} \\
& \text { Let } \frac{P(x)}{Q(x)}=\frac{A}{\left(x-\alpha_1\right)}+\frac{B}{\left(x-\alpha_2\right)} \cdots
\end{aligned}
$

Find $A, B$...
By comparing $N^x$ and $P(x)$

$I=\int \frac{6 d x}{(x-1)(x+1)}=\int\left(\frac{A}{(x-1)}+\frac{B}{(x+1)}\right) d x$
Thus $6=A(x+1)+B(x-1)$
On calculating $A=3, B=-3$
Thus $I=3 \ln (x-1)-3 \ln (x+1)+C$
Hence, the answer is the option (3).

Example 5: Find $\int \frac{d x}{x^2-4}$ :
1) $\ln \frac{x+2}{x-2}+C$
2) $1 / 2 \ln \frac{x-2}{x+2}+C$
3) $\frac{1}{4} \ln \left|\frac{x-2}{x+2}\right|+C$
4) none of these

Solution

As we have learned

Rule of integration by Partial fraction -

Linear and non-repeated:
$\frac{P(x)}{Q(x)}=\frac{P(x)}{\left(x-\alpha_1\right)\left(x-\alpha_2\right) \cdots\left(x-\alpha_n\right)}$
Let $\frac{P(x)}{Q(x)}=\frac{A}{\left(x-\alpha_1\right)}+\frac{B}{\left(x-\alpha_2\right)} \cdots$
Find $A, B$...
By comparing $N^x$ and $P(x)$

$\begin{aligned}
& I=\int \frac{d x}{x^2-4}=\int \frac{d x}{(x-2)(x+2)} \\
& =-\frac{\ln (|x+2|)-\ln (|x-2|)}{4}+C \\
& =\frac{1}{4} \ln \left|\frac{x-2}{x+2}\right|+C
\end{aligned}$
Hence, the answer is the option (3).

List of Topics Related to Integration by Partial fractions

This section highlights related topics that strengthen your understanding of integration as a whole-from basic indefinite integrals to specific applications in inequalities and irrational functions. Explore each link to see how these concepts build upon one another to form a complete picture of integral calculus.

Application of Integrals

Integral of Particular Functions

Indefinite Integrals

Integration of irrational functions

Application of Inequality in Definite Integration

NCERT Resources

This section gathers all the essential NCERT study materials to help you master Chapter 7 – Integrals. From concise notes to detailed solutions and exemplar problems, it provides everything you need in one place for organized and effective exam preparation.

NCERT Class 12 Maths Notes for Chapter 7 - Integrals

NCERT Class 12 Maths Solutions for Chapter 7 - Integrals

NCERT Class 12 Maths Exemplar Solutions for Chapter 7 - Integrals

Practice Questions based on Integration by Partial fractions

This section offers practice questions focused on Integration by Partial Fractions to help reinforce your understanding and application of the method. These problems provide valuable hands-on experience, boosting your confidence in solving various integration concepts effectively.

Integration Using Partial Fraction- Practice Question MCQ

We have provided below the practice questions related to different concepts of integration to improve your understanding: