Integration by Partial Fractions

Q: What is the method of partial fractions?

A technique for decomposing a proper rational function P(x)/Q(x) into a sum of simpler fractions - each involving one linear or one irreducible quadratic factor of Q(x). Once decomposed, each piece integrates to a logarithm or arctangent.

Q: When can I use partial fractions?

Any rational function - once made proper by long division if needed - can be decomposed. It is the standard route for integrating P(x)/Q(x) when direct substitution or by-parts do not apply. Also useful after a substitution reduces a trig or radical integral to a rational one.

Q: How do I find the constants A, B in the decomposition?

Two methods: comparison of coefficients (multiply both sides by Q(x) and equate powers of x, solve the linear system) or numerical substitution (plug in convenient values of x, usually roots of Q). For distinct linear factors, substitution is faster; for quadratic factors, comparison or a mix is needed.

Q: What if Q(x) has an irreducible quadratic factor?

Assign a numerator of the form Mx + N over that factor, not just a constant A. Determine M and N by comparison of coefficients. To integrate (Mx + N)/(x^2 + alpha x + beta), split the numerator into a multiple of the derivative of the denominator plus a constant.

Q: What if Q(x) has a repeated linear factor?

For (x - a)^k, include every power from 1 to k: A_1/(x-a) + A_2/(x-a)^2 + ... + A_k/(x-a)^k. Each subsequent power integrates as a negative reciprocal, and the first-power piece integrates as ln|x - a|.

Q: Can partial fractions be used with trigonometric integrals?

Yes - very often via the Weierstrass substitution. For a rational function of sin x and cos x, put t = tan(x/2) to convert it into a rational function of t, then apply partial fractions.

Q: What if the degree of P(x) is at least the degree of Q(x)?

Do polynomial long division first, writing P(x)/Q(x) = h(x) + p(x)/Q(x), where h(x) is a polynomial and p(x) has degree less than Q. Integrate h(x) as a polynomial and apply partial fractions to p(x)/Q(x).

Q: Does every rational function integrate to elementary functions?

Yes. Every rational function has an elementary antiderivative - a sum of rational functions, logarithms, and arctangents. This is a classical result of calculus and is why partial fractions is such a powerful method: it is guaranteed to give a closed-form answer.

MathsIntegralsFor JEE aspirants

Integration by partial fractions is the technique for integrating a rational function $\frac{P ( x )}{Q ( x )}$ by first decomposing it into a sum of simpler fractions, each of which is a standard integral. For a proper rational fraction (degree of $P$ less than degree of $Q$ ), factor $Q (x)$ into distinct linear factors $(x - a)$ , repeated linear factors $(x - a)^{k}$ , and irreducible quadratic factors $(x^{2} + α x + β)$ . Assign the appropriate partial-fraction template, solve for the constants by equating coefficients (or by clever substitution), and integrate each fraction using standard formulae. This technique is essential for JEE Maths and often appears combined with Weierstrass substitution for trigonometric rational integrals.

Key Formulas - Quick Reference

$\int \frac{d x}{x - a} = ln ∣ x - a ∣ + C$
$\int \frac{d x}{( x - a ) ^{k}} = \frac{- 1}{( k - 1 ) ( x - a ) ^{k - 1}} + C, k > 1$
$\int \frac{d x}{x ^{2} + a ^{2}} = \frac{1}{a} tan^{- 1} \frac{x}{a} + C$
Distinct linear factors: $\frac{P ( x )}{( x - a ) ( x - b )} = \frac{A}{x - a} + \frac{B}{x - b}$
Repeated linear factor $(x - a)^{k}$ : $\frac{A _{1}}{x - a} + \frac{A _{2}}{( x - a ) ^{2}} + \dots + \frac{A _{k}}{( x - a ) ^{k}}$
Irreducible quadratic $(x^{2} + α x + β)^{r}$ : $\frac{M _{1} x + N _{1}}{x ^{2} + α x + β} + \dots + \frac{M _{r} x + N _{r}}{( x ^{2} + α x + β ) ^{r}}$

1. What is a Partial Fraction Decomposition?

A rational function is a ratio $\frac{P ( x )}{Q ( x )}$ of two polynomials. The idea of partial fractions is to write this single fraction as a sum of simpler fractions - each of which is a standard integral.

Example:

$\frac{x + 7}{( 2 x - 3 ) ( 3 x + 4 )} = \frac{1}{2 x - 3} - \frac{1}{3 x + 4} .$

The RHS is called the partial-fraction decomposition of the LHS. Once decomposed, each piece integrates as a logarithm.

Proper vs improper fractions

A rational function $\frac{P ( x )}{Q ( x )}$ is proper if $de g P < de g Q$ ; otherwise it is improper.

If improper, first perform polynomial long division to write $\frac{P ( x )}{Q ( x )} = h (x) + \frac{p ( x )}{Q ( x )}$ , where $h (x)$ is a polynomial (easy to integrate) and $\frac{p ( x )}{Q ( x )}$ is proper (decompose it).

2. The Decomposition Template

Factor $Q (x)$ completely over the reals:

$Q (x) = (x - a)^{k} \dots (x^{2} + α x + β)^{r} \dots$

Then set up the partial-fraction template:

$\frac{P ( x )}{Q ( x )} = one group per linear factor of multiplicity k \frac{A _{1}}{x - a} + \frac{A _{2}}{( x - a ) ^{2}} + \dots + \frac{A _{k}}{( x - a ) ^{k}}$

$+ one group per irreducible quadratic factor of multiplicity r \frac{M _{1} x + N _{1}}{x ^{2} + α x + β} + \frac{M _{2} x + N _{2}}{( x ^{2} + α x + β ) ^{2}} + \dots + \frac{M _{r} x + N _{r}}{( x ^{2} + α x + β ) ^{r}} + \dots$

The unknowns $A_{i}, M_{j}, N_{j}$ are constants. Determine them by either:

Comparison of coefficients: multiply both sides by $Q (x)$ ; equate coefficients of like powers of $x$ to get a linear system.
Numerical substitution: plug in clever values of $x$ (typically the roots of $Q (x)$ ) to solve for individual constants directly.

Quick reference: template by denominator shape

Factor in $Q (x)$	Partial-fraction contribution
$(x - a)$ - distinct linear	$\frac{A}{x - a}$
$(x - a)^{k}$ - repeated linear	$\frac{A _{1}}{x - a} + \frac{A _{2}}{( x - a ) ^{2}} + \dots + \frac{A _{k}}{( x - a ) ^{k}}$
$(x^{2} + α x + β)$ - irreducible quadratic ( $α^{2} < 4 β$ )	$\frac{M x + N}{x ^{2} + α x + β}$
$(x^{2} + α x + β)^{r}$ - repeated irreducible quadratic	$\frac{M _{1} x + N _{1}}{x ^{2} + α x + β} + \dots + \frac{M _{r} x + N _{r}}{( x ^{2} + α x + β ) ^{r}}$

3. Worked Examples

Solved Example 1

Evaluate

\int \frac{cos x d x}{( 1 + sin x ) ( 2 + sin x )} .

Solution:

Put $sin x = t$ , so $cos x d x = d t$ :

$I = \int \frac{d t}{( 1 + t ) ( 2 + t )} .$

Decompose: $\frac{1}{( 1 + t ) ( 2 + t )} = \frac{A}{1 + t} + \frac{B}{2 + t}$ .

Multiplying both sides by $(1 + t) (2 + t)$ : $1 = A (2 + t) + B (1 + t)$ . Setting $t = - 1$ : $1 = A (1) \Rightarrow A = 1$ . Setting $t = - 2$ : $1 = B (- 1) \Rightarrow B = - 1$ .

$I = \int \frac{d t}{1 + t} - \int \frac{d t}{2 + t} = ln ∣1 + t ∣ - ln ∣2 + t ∣ + C = ln \frac{1 + sin x}{2 + sin x} + C .$

Solved Example 2

Evaluate

\int \frac{x d x}{x ^{3} + 1} .

Solution:

Factor: $x^{3} + 1 = (x + 1) (x^{2} - x + 1)$ . The quadratic factor is irreducible (discriminant $1 - 4 = - 3 < 0$ ). So

$\frac{x}{x ^{3} + 1} = \frac{A}{x + 1} + \frac{B x + D}{x ^{2} - x + 1} .$

Multiplying by $x^{3} + 1$ : $x = A (x^{2} - x + 1) + (B x + D) (x + 1)$ .

Expanding: $x = (A + B) x^{2} + (- A + B + D) x + (A + D)$ . Equating coefficients:

$x^{2}$ : $A + B = 0$
$x$ : $- A + B + D = 1$
const: $A + D = 0$

Solving: $A = - \frac{1}{3}, B = \frac{1}{3}, D = \frac{1}{3}$ .

$I = - \frac{1}{3} \int \frac{d x}{x + 1} + \frac{1}{3} \int \frac{x + 1}{x ^{2} - x + 1} d x .$

The first is $- \frac{1}{3} ln ∣ x + 1∣$ . For the second, write $x + 1 = \frac{1}{2} (2 x - 1) + \frac{3}{2}$ :

$\int \frac{x + 1}{x ^{2} - x + 1} d x = \frac{1}{2} \int \frac{2 x - 1}{x ^{2} - x + 1} d x + \frac{3}{2} \int \frac{d x}{x ^{2} - x + 1} .$

First piece: $\frac{1}{2} ln ∣ x^{2} - x + 1∣$ (numerator is exactly the derivative of the denominator).

Second piece: complete the square, $x^{2} - x + 1 = (x - \frac{1}{2})^{2} + \frac{3}{4}$ . Then

$\frac{3}{2} \int \frac{d x}{( x - \frac{1}{2} ) ^{2} + ( \frac{3}{2} ) ^{2}} = \frac{3}{2} \cdot \frac{2}{3} tan^{- 1} (\frac{2 x - 1}{3}) = 3 tan^{- 1} (\frac{2 x - 1}{3}) .$

Combining everything and multiplying by $\frac{1}{3}$ :

$I = - \frac{1}{3} ln ∣ x + 1∣ + \frac{1}{6} ln ∣ x^{2} - x + 1∣ + \frac{1}{3} tan^{- 1} (\frac{2 x - 1}{3}) + C .$

Solved Example 3

Evaluate

\int \frac{d x}{sin x ( 2 + cos x - 2 sin x )} .

Solution:

Apply the Weierstrass substitution $t = tan (x /2)$ : $sin x = \frac{2 t}{1 + t ^{2}}, cos x = \frac{1 - t ^{2}}{1 + t ^{2}}, d x = \frac{2 d t}{1 + t ^{2}}$ .

Substituting:

$I = \int \frac{\frac{2 d t}{1 + t ^{2}}}{\frac{2 t}{1 + t ^{2}} ( 2 + \frac{1 - t ^{2}}{1 + t ^{2}} - \frac{4 t}{1 + t ^{2}} )} .$

Simplify the parenthesis: $2 + \frac{1 - t ^{2} - 4 t}{1 + t ^{2}} = \frac{2 ( 1 + t ^{2} ) + 1 - t ^{2} - 4 t}{1 + t ^{2}} = \frac{t ^{2} - 4 t + 3}{1 + t ^{2}}$ .

$I = \int \frac{\frac{2 d t}{1 + t ^{2}}}{\frac{2 t ( t ^{2} - 4 t + 3 )}{( 1 + t ^{2} ) ^{2}}} = \int \frac{( 1 + t ^{2} ) d t}{t ( t ^{2} - 4 t + 3 )} = \int \frac{( 1 + t ^{2} ) d t}{t ( t - 1 ) ( t - 3 )} .$

Partial fractions:

$\frac{1 + t ^{2}}{t ( t - 1 ) ( t - 3 )} = \frac{A}{t} + \frac{B}{t - 3} + \frac{C}{t - 1} .$

Multiplying and substituting $t = 0$ : $1 = A (- 1) (- 3) = 3 A \Rightarrow A = \frac{1}{3}$ .
$t = 3$ : $10 = B (3) (2) = 6 B \Rightarrow B = \frac{5}{3}$ .
$t = 1$ : $2 = C (1) (- 2) = - 2 C \Rightarrow C = - 1$ .

$I = \frac{1}{3} \int \frac{d t}{t} + \frac{5}{3} \int \frac{d t}{t - 3} - \int \frac{d t}{t - 1} = \frac{1}{3} ln ∣ t ∣ + \frac{5}{3} ln ∣ t - 3∣ - ln ∣ t - 1∣ + C .$

Back-substitute $t = tan (x /2)$ :

$I = \frac{1}{3} ln tan \frac{x}{2} + \frac{5}{3} ln tan \frac{x}{2} - 3 - ln tan \frac{x}{2} - 1 + C .$

Common Mistakes to Avoid

Watch out

Not checking if the fraction is proper first. If $de g P \geq de g Q$ , you must divide first. Attempting a partial-fraction decomposition on an improper fraction gives a linear system with no solution.
Missing the $\frac{M x + N}{\dots}$ form for irreducible quadratic denominators. The numerator over $(x^{2} + α x + β)$ must be a general linear $M x + N$ , not a constant. Using just $A$ instead loses a degree of freedom and the system becomes inconsistent.
Only writing $\frac{A}{( x - a ) ^{k}}$ for repeated linear factors. A factor $(x - a)^{k}$ requires all powers from 1 to $k$ : $\frac{A _{1}}{x - a} + \frac{A _{2}}{( x - a ) ^{2}} + \dots + \frac{A _{k}}{( x - a ) ^{k}}$ . Skipping intermediate powers is a common error.
Using the substitution shortcut with irreducible quadratic factors. The "plug in the root" trick works cleanly for distinct linear factors (real roots). For quadratic factors, you have to use comparison of coefficients (or a mix).
Absolute value in $ln$ . After decomposing and integrating, always write $ln ∣ x - a ∣$ , not $ln (x - a)$ . This matters when the sign of $x - a$ can be negative in the problem's domain.

Frequently Asked Questions

Q1. What is the method of partial fractions?

A technique for decomposing a rational function $\frac{P ( x )}{Q ( x )}$ (with $de g P < de g Q$ ) into a sum of simpler fractions - each involving one linear or one irreducible quadratic factor of $Q (x)$ . Once decomposed, each piece integrates to a logarithm or arctangent.

Q2. When can I use partial fractions?

Any rational function - once ensured proper by long division if needed - can be decomposed. It's the standard route for integrals like $\int \frac{P ( x )}{Q ( x )} d x$ where direct substitution or by-parts don't apply. Also useful after a substitution reduces a trigonometric or radical integral to a rational one.

Q3. How do I find the constants $A, B, \dots$ in the decomposition?

Two methods: (1) Comparison of coefficients - multiply both sides by $Q (x)$ and equate the coefficient of each power of $x$ ; solve the resulting linear system. (2) Numerical substitution - plug in convenient values of $x$ (usually the roots of $Q$ ) to isolate individual constants directly. For distinct linear factors, method (2) is faster; for quadratic factors, method (1) or a mix is needed.

Q4. What if $Q (x)$ has an irreducible quadratic factor?

Assign a numerator of the form $M x + N$ over that factor (not just a constant $A$ ). Determine $M$ and $N$ by comparison of coefficients. To integrate $\int \frac{M x + N}{x ^{2} + α x + β} d x$ , split the numerator as $\frac{M}{2} (2 x + α) + constant$ , then use $\int f ^{'} / f$ and $\int \frac{d x}{u ^{2} + a ^{2}}$ formulae.

Q5. What if $Q (x)$ has a repeated linear factor?

For $(x - a)^{k}$ , include every power from 1 to $k$ : $\frac{A _{1}}{x - a} + \frac{A _{2}}{( x - a ) ^{2}} + \dots + \frac{A _{k}}{( x - a ) ^{k}}$ . Each subsequent power integrates as $\frac{- 1}{( n - 1 ) ( x - a ) ^{n - 1}}$ (or $ln ∣ x - a ∣$ for the first-power piece).

Q6. Can partial fractions be used with trigonometric integrals?

Yes - very often via the Weierstrass substitution. If you have a rational function of $sin x$ and $cos x$ , put $t = tan (x /2)$ to convert it into a rational function of $t$ , then apply partial fractions. Solved Example 3 shows this workflow end-to-end.

Q7. What if the degree of $P (x)$ is at least the degree of $Q (x)$ ?

Do polynomial long division first, writing $\frac{P ( x )}{Q ( x )} = h (x) + \frac{p ( x )}{Q ( x )}$ , where $h (x)$ is a polynomial and $p (x)$ has $de g p < de g Q$ . Integrate $h (x)$ as a polynomial and apply partial fractions to $\frac{p ( x )}{Q ( x )}$ .

Q8. Does every rational function integrate to elementary functions?

Yes. Every rational function has an elementary antiderivative - a sum of rational functions, logarithms, and arctangents. This is one of the classical results of calculus and is why partial fractions is such a powerful method: it's guaranteed to give a closed-form answer.

Previous year questions on Integration by Partial Fractions

1 question from past papers, each with a step-by-step solution.

JEE Main 2026 Apr 2 Shift 2, Mathematics Q19

Ready to master Integrals?

Take a full mock test, practice concept-by-concept, and get an AI-powered rank prediction — all on Fundamenthol.

Practice questions →Take a mock test

1. What is a Partial Fraction Decomposition?

Proper vs improper fractions

2. The Decomposition Template

Quick reference: template by denominator shape

3. Worked Examples

Common Mistakes to Avoid

Frequently Asked Questions

Q1. What is the method of partial fractions?

Q2. When can I use partial fractions?

Q3. How do I find the constants A,B,… in the decomposition?

Q4. What if Q(x) has an irreducible quadratic factor?

Q5. What if Q(x) has a repeated linear factor?

Q6. Can partial fractions be used with trigonometric integrals?

Q7. What if the degree of P(x) is at least the degree of Q(x)?

Q8. Does every rational function integrate to elementary functions?

Previous year questions on Integration by Partial Fractions

Ready to master Integrals?

Q3. How do I find the constants $A, B, \dots$ in the decomposition?

Q4. What if $Q (x)$ has an irreducible quadratic factor?

Q5. What if $Q (x)$ has a repeated linear factor?

Q7. What if the degree of $P (x)$ is at least the degree of $Q (x)$ ?