Partial Fractions Integration — Edexcel A-level Mathematics (Pure)

SPEC 8.12 • INTEGRATION

Partial Fractions for Integration

Decomposing rational functions to find their integrals using logarithms

Why Partial Fractions?

Many rational functions cannot be integrated directly. Partial fractions breaks them into simpler pieces that we know how to integrate.

Key Idea

A complex fraction like

\frac{1}{( x - 1 ) ( x + 2 )}

cannot be integrated directly.

But

\frac{1}{3 ( x - 1 )} - \frac{1}{3 ( x + 2 )}

gives us two simple logarithms!

Prerequisites

You should already know how to:

Decompose fractions into partial fractions (from Pure Maths)

Integrate

\frac{1}{x - a}

to get

ln ∣ x - a ∣ + C

Type 1: Distinct Linear Factors

When the denominator has different linear factors:

Form

\frac{f ( x )}{( x - a ) ( x - b )} = \frac{A}{x - a} + \frac{B}{x - b}

Integration result:

Formula

\int \frac{A}{x - a} + \frac{B}{x - b} d x = A ln ∣ x - a ∣ + B ln ∣ x - b ∣ + C

Tip

For three or more distinct factors, simply add more terms. Each linear factor

(x - c)

contributes a term

\frac{C}{x - c}

.

Worked Example01

Find $\int \frac{1}{( x - 1 ) ( x + 2 )} d x$

1

Set up partial fractions:

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

2

Multiply through:

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

3

Let x = 1:

1 = A (3)

, so

A = \frac{1}{3}

4

Let x = -2:

1 = B (- 3)

, so

B = - \frac{1}{3}

5

So:

\int \frac{1}{( x - 1 ) ( x + 2 )} d x = \int \frac{1/3}{x - 1} - \frac{1/3}{x + 2} d x

6

Integrate:

= \frac{1}{3} ln ∣ x - 1∣ - \frac{1}{3} ln ∣ x + 2∣ + C

Answer

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

Worked Example02

Find $\int \frac{2 x + 5}{( x + 1 ) ( x - 1 ) ( x + 2 )} d x$

1

Set up:

\frac{2 x + 5}{( x + 1 ) ( x - 1 ) ( x + 2 )} = \frac{A}{x + 1} + \frac{B}{x - 1} + \frac{C}{x + 2}

2

Multiply through:

2 x + 5 = A (x - 1) (x + 2) + B (x + 1) (x + 2) + C (x + 1) (x - 1)

3

Let x = -1:

3 = A (- 2) (1) = - 2 A

, so

A = - \frac{3}{2}

4

Let x = 1:

7 = B (2) (3) = 6 B

, so

B = \frac{7}{6}

5

Let x = -2:

1 = C (- 1) (- 3) = 3 C

, so

C = \frac{1}{3}

6

Integrate:

\int (- \frac{3}{2} \cdot \frac{1}{x + 1} + \frac{7}{6} \cdot \frac{1}{x - 1} + \frac{1}{3} \cdot \frac{1}{x + 2}) d x

Answer

- \frac{3}{2} ln ∣ x + 1∣ + \frac{7}{6} ln ∣ x - 1∣ + \frac{1}{3} ln ∣ x + 2∣ + C

Type 2: Repeated Linear Factors

When a factor appears more than once, we need additional terms:

Form for (x-a)²

\frac{f ( x )}{( x - a ) ^{2} ( x - b )} = \frac{A}{x - a} + \frac{B}{( x - a ) ^{2}} + \frac{C}{x - b}

Key integrals for repeated factors:

IntegralResult

\int \frac{1}{( x - a ) ^{2}} d x

- \frac{1}{x - a} + C

\int \frac{1}{( x - a ) ^{3}} d x

- \frac{1}{2 ( x - a ) ^{2}} + C

\int \frac{1}{( x - a ) ^{n}} d x

- \frac{1}{( n - 1 ) ( x - a ) ^{n - 1}} + C

Worked Example03

Find $\int \frac{3 x + 1}{( x - 2 ) ^{2} ( x + 1 )} d x$

1

Set up:

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

2

Multiply through:

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

3

Let x = 2:

7 = B (3)

, so

B = \frac{7}{3}

4

Let x = -1:

- 2 = C (9)

, so

C = - \frac{2}{9}

5

Compare x² coefficients:

0 = A + C

, so

A = \frac{2}{9}

6

Integrate:

\int \frac{2/9}{x - 2} + \frac{7/3}{( x - 2 ) ^{2}} - \frac{2/9}{x + 1} d x

Answer

\frac{2}{9} ln ∣ x - 2∣ - \frac{7}{3 ( x - 2 )} - \frac{2}{9} ln ∣ x + 1∣ + C

Worked Example04

Evaluate $\int_{0}^{1} \frac{x ^{2}}{( x + 1 ) ^{3}} d x$

1

Set up:

\frac{x ^{2}}{( x + 1 ) ^{3}} = \frac{A}{x + 1} + \frac{B}{( x + 1 ) ^{2}} + \frac{C}{( x + 1 ) ^{3}}

2

Multiply through:

x^{2} = A (x + 1)^{2} + B (x + 1) + C

3

Let x = -1:

1 = C

. Compare x²:

A = 1

. Compare x:

0 = 2 A + B

, so

B = - 2

4

Integrate:

[ln ∣ x + 1∣ + \frac{2}{x + 1} - \frac{1}{2 ( x + 1 ) ^{2}}]_{0}^{1}

5

At x = 1:

ln 2 + 1 - \frac{1}{8}

. At x = 0:

ln 1 + 2 - \frac{1}{2} = \frac{3}{2}

6

Subtract:

ln 2 + 1 - \frac{1}{8} - \frac{3}{2} = ln 2 - \frac{5}{8}

Answer

ln 2 - \frac{5}{8}

Improper Fractions

When the degree of the numerator is ≥ the degree of the denominator, we must do polynomial division first.

Method

Step 1: Divide numerator by denominator to get quotient + remainder/denominator

Step 2: Apply partial fractions to the remainder term

Step 3: Integrate each part

Example

\frac{x ^{3}}{( x - 1 ) ( x + 1 )}

is improper (degree 3 ≥ degree 2)

First divide:

\frac{x ^{3}}{x ^{2} - 1} = x + \frac{x}{x ^{2} - 1}

Then partial fractions on

\frac{x}{x ^{2} - 1}

Common Error

Forgetting to check if the fraction is improper before applying partial fractions!

Worked Example05

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

1

This is improper (degree 3 > degree 2). Divide first.

2

x^{3} + 2 x = (x^{2} - 1) (x) + 3 x

, so

\frac{x ^{3} + 2 x}{x ^{2} - 1} = x + \frac{3 x}{x ^{2} - 1}

3

Partial fractions:

\frac{3 x}{( x - 1 ) ( x + 1 )} = \frac{A}{x - 1} + \frac{B}{x + 1}

4

Let x = 1:

3 = 2 A

, so

A = \frac{3}{2}

. Let x = -1:

- 3 = - 2 B

, so

B = \frac{3}{2}

5

Integrate:

\int x + \frac{3/2}{x - 1} + \frac{3/2}{x + 1} d x

Answer

\frac{x ^{2}}{2} + \frac{3}{2} ln ∣ x - 1∣ + \frac{3}{2} ln ∣ x + 1∣ + C = \frac{x ^{2}}{2} + \frac{3}{2} ln ∣ x^{2} - 1∣ + C

Worked Example06

Find $\int \frac{x ^{4} + x ^{3} - x + 2}{x ^{2} + x - 2} d x$

1

Improper (degree 4 > degree 2). Divide:

x^{4} + x^{3} - x + 2

by

x^{2} + x - 2

2

Long division gives:

x^{2} + 2 + \frac{3 x - 2}{x ^{2} + x - 2}

3

Factorise:

x^{2} + x - 2 = (x + 2) (x - 1)

4

Partial fractions:

\frac{3 x - 2}{( x + 2 ) ( x - 1 )} = \frac{A}{x + 2} + \frac{B}{x - 1}

5

Let x = 1:

1 = 3 B

, so

B = \frac{1}{3}

. Let x = -2:

- 8 = - 3 A

, so

A = \frac{8}{3}

6

Integrate:

\int x^{2} + 2 + \frac{8/3}{x + 2} + \frac{1/3}{x - 1} d x

Answer

\frac{x ^{3}}{3} + 2 x + \frac{8}{3} ln ∣ x + 2∣ + \frac{1}{3} ln ∣ x - 1∣ + C

Practice

Question 1: Find

\int \frac{5}{( x - 3 ) ( x + 2 )} d x

Set up:

\frac{5}{( x - 3 ) ( x + 2 )} = \frac{A}{x - 3} + \frac{B}{x + 2}

Let x = 3:

5 = 5 A

, so A = 1. Let x = -2:

5 = - 5 B

, so B = -1.

\int \frac{1}{x - 3} - \frac{1}{x + 2} d x = ln ∣ x - 3∣ - ln ∣ x + 2∣ + C

Answer

ln \frac{x - 3}{x + 2} + C

Practice

Question 2: Find

\int \frac{2 x + 3}{( x + 1 ) ^{2}} d x

Set up:

\frac{2 x + 3}{( x + 1 ) ^{2}} = \frac{A}{x + 1} + \frac{B}{( x + 1 ) ^{2}}

2 x + 3 = A (x + 1) + B

. Let x = -1: B = 1. Compare x: A = 2.

\int \frac{2}{x + 1} + \frac{1}{( x + 1 ) ^{2}} d x = 2 ln ∣ x + 1∣ - \frac{1}{x + 1} + C

Answer

2 ln ∣ x + 1∣ - \frac{1}{x + 1} + C

Practice

Question 3: Find

\int \frac{x ^{2} + 3}{x ^{2} - 4} d x

Improper fraction. Divide:

\frac{x ^{2} + 3}{x ^{2} - 4} = 1 + \frac{7}{x ^{2} - 4}

Partial fractions:

\frac{7}{( x - 2 ) ( x + 2 )} = \frac{A}{x - 2} + \frac{B}{x + 2}

Let x = 2:

7 = 4 A

, so

A = \frac{7}{4}

. Let x = -2:

7 = - 4 B

, so

B = - \frac{7}{4}

\int 1 + \frac{7/4}{x - 2} - \frac{7/4}{x + 2} d x

Answer

x + \frac{7}{4} ln \frac{x - 2}{x + 2} + C

Summary

Key Integration Results

FormIntegral

\frac{A}{x - a}

A ln ∣ x - a ∣ + C

\frac{B}{( x - a ) ^{2}}

- \frac{B}{x - a} + C

\frac{B}{( x - a ) ^{n}}

- \frac{B}{( n - 1 ) ( x - a ) ^{n - 1}} + C

Exam Tips

Always check if the fraction is improper first - if so, divide!

Show your partial fraction decomposition clearly

Combine logarithms for neat final answers:

ln a - ln b = ln \frac{a}{b}

Remember: repeated factors need multiple terms