Partial Fractions

Decomposing algebraic fractions into simpler components for integration and analysis

SPEC 2.10 • ALGEBRA AND FUNCTIONS

What Are Partial Fractions?

Partial fractions is the reverse process of adding algebraic fractions together.

Key Concept

We can combine fractions:

\frac{2}{x + 1} + \frac{3}{x - 2} = \frac{5 x - 1}{( x + 1 ) ( x - 2 )}

Partial fractions reverses this:

\frac{5 x - 1}{( x + 1 ) ( x - 2 )} = \frac{2}{x + 1} + \frac{3}{x - 2}

Why Use Partial Fractions?

Integration: Simpler fractions are much easier to integrate

Series expansions: Required for binomial expansions of rational functions

Differential equations: Used in solving certain types of ODEs

Essential First Step

The denominator must be fully factorised before you can decompose into partial fractions.

The Two Types of Partial Fractions

The form of your partial fractions depends on the factors in the denominator:

Type 1: Distinct Linear Factors

\frac{p x + q}{( a x + b ) ( c x + d )}

becomes

\frac{A}{a x + b} + \frac{B}{c x + d}

Each different linear factor gets its own fraction with a constant numerator.

Type 2: Repeated Linear Factors

\frac{p x + q}{( a x + b ) ^{2}}

becomes

\frac{A}{a x + b} + \frac{B}{( a x + b ) ^{2}}

A squared factor needs TWO terms: one for each power.

Common Error

For repeated factors like

(x - 1)^{2}

, you need BOTH

\frac{A}{x - 1}

AND

\frac{B}{( x - 1 ) ^{2}}

, not just the squared term!

Proper vs Improper Fractions

Critical Rule

Partial fractions only work directly on PROPER fractions

A fraction is proper if the degree of the numerator is LESS than the degree of the denominator.

FractionNumerator DegreeDenominator DegreeType

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

12Proper

\frac{x ^{2} + 1}{x ^{2} - 4}

22Improper

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

32Improper

For Improper Fractions

You must use algebraic division first to get a polynomial plus a proper fraction, then apply partial fractions to the proper fraction part.

Finding the Constants: Two Methods

Method 1: Substitution

1

Multiply both sides by the denominator

2

Substitute values of x that make factors zero

3

Each substitution eliminates all but one unknown

Best for: Finding constants quickly when you have linear factors

Method 2: Comparing Coefficients

1

Multiply both sides by the denominator

2

Expand the right-hand side

3

Compare coefficients of like powers of x

4

Solve simultaneous equations

Best for: Finding remaining constants or checking your answer

Pro Tip

Often the fastest approach is to use substitution to find as many constants as possible, then use comparing coefficients (or x = 0) to find any remaining constants.

Worked Example - Distinct Linear Factors (Substitution)01

Express $\frac{7 x - 1}{( x + 1 ) ( x - 2 )}$ in partial fractions.

1

Write in partial fraction form:

\frac{7 x - 1}{( x + 1 ) ( x - 2 )} = \frac{A}{x + 1} + \frac{B}{x - 2}

2

Multiply both sides by

(x + 1) (x - 2)

:

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

3

Substitute

x = 2

:

7 (2) - 1 = A (0) + B (3)

so

13 = 3 B

giving

B = \frac{13}{3}

4

Substitute

x = - 1

:

7 (- 1) - 1 = A (- 3) + B (0)

so

- 8 = - 3 A

giving

A = \frac{8}{3}

Answer

\frac{7 x - 1}{( x + 1 ) ( x - 2 )} = \frac{8}{3 ( x + 1 )} + \frac{13}{3 ( x - 2 )}

Worked Example - Distinct Linear Factors (Comparing Coefficients)02

Express $\frac{5 x + 7}{( x + 2 ) ( x + 3 )}$ in partial fractions using comparing coefficients.

1

Write:

\frac{5 x + 7}{( x + 2 ) ( x + 3 )} = \frac{A}{x + 2} + \frac{B}{x + 3}

2

Multiply through:

5 x + 7 = A (x + 3) + B (x + 2)

3

Expand:

5 x + 7 = A x + 3 A + B x + 2 B = (A + B) x + (3 A + 2 B)

4

Compare coefficients of x:

A + B = 5

5

Compare constant terms:

3 A + 2 B = 7

6

Solve: From (4),

A = 5 - B

. Sub into (5):

3 (5 - B) + 2 B = 7

so

15 - B = 7

giving

B = 8

,

A = - 3

Answer

\frac{5 x + 7}{( x + 2 ) ( x + 3 )} = \frac{- 3}{x + 2} + \frac{8}{x + 3}

Worked Example - Three Distinct Linear Factors03

Express $\frac{2 x ^{2} + 5 x - 10}{( x - 1 ) ( x + 1 ) ( x + 2 )}$ in partial fractions.

1

Write:

\frac{2 x ^{2} + 5 x - 10}{( x - 1 ) ( x + 1 ) ( x + 2 )} = \frac{A}{x - 1} + \frac{B}{x + 1} + \frac{C}{x + 2}

2

Multiply through:

2 x^{2} + 5 x - 10 = A (x + 1) (x + 2) + B (x - 1) (x + 2) + C (x - 1) (x + 1)

3

Let

x = 1

:

2 + 5 - 10 = A (2) (3)

so

- 3 = 6 A

giving

A = - \frac{1}{2}

4

Let

x = - 1

:

2 - 5 - 10 = B (- 2) (1)

so

- 13 = - 2 B

giving

B = \frac{13}{2}

5

Let

x = - 2

:

8 - 10 - 10 = C (- 3) (- 1)

so

- 12 = 3 C

giving

C = - 4

Answer

\frac{2 x ^{2} + 5 x - 10}{( x - 1 ) ( x + 1 ) ( x + 2 )} = \frac{- 1}{2 ( x - 1 )} + \frac{13}{2 ( x + 1 )} - \frac{4}{x + 2}

Repeated Linear Factors

When a factor appears more than once (squared), you need a term for each power of that factor.

Form for Repeated Factors

For

(a x + b)^{2}

in the denominator, include:

\frac{A}{a x + b} + \frac{B}{( a x + b ) ^{2}}

You need BOTH terms, not just the squared term.

Mixed Example

For

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

, you would write:

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

Strategy

Use substitution to find the constant for the highest power first (it's the only one that survives when you set x equal to the root), then use another substitution or comparing coefficients for the remaining constants.

Worked Example - Repeated Linear Factor04

Express $\frac{3 x + 5}{( x + 1 ) ( x - 1 ) ^{2}}$ in partial fractions.

1

Write:

\frac{3 x + 5}{( x + 1 ) ( x - 1 ) ^{2}} = \frac{A}{x + 1} + \frac{B}{x - 1} + \frac{C}{( x - 1 ) ^{2}}

2

Multiply through:

3 x + 5 = A (x - 1)^{2} + B (x + 1) (x - 1) + C (x + 1)

3

Let

x = 1

:

3 + 5 = C (2)

so

8 = 2 C

giving

C = 4

4

Let

x = - 1

:

- 3 + 5 = A (- 2)^{2}

so

2 = 4 A

giving

A = \frac{1}{2}

5

Let

x = 0

:

5 = A (1) + B (- 1) + C (1) = \frac{1}{2} - B + 4

so

B = - \frac{1}{2}

Answer

\frac{3 x + 5}{( x + 1 ) ( x - 1 ) ^{2}} = \frac{1}{2 ( x + 1 )} - \frac{1}{2 ( x - 1 )} + \frac{4}{( x - 1 ) ^{2}}

Improper Fractions

When the degree of the numerator is greater than or equal to the degree of the denominator, you must use algebraic division first.

Process for Improper Fractions

Step 1: Perform algebraic division to get: Quotient +

\frac{Remainder}{Divisor}

Step 2: Apply partial fractions to the proper fraction (the remainder part)

Step 3: Combine the quotient with your partial fractions

Example Structure

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

is improper (degree 3 vs degree 2)

After division:

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

Then apply partial fractions to

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

Worked Example - Improper Fraction05

Express $\frac{x ^{3} + x ^{2} - 2}{( x - 1 ) ( x + 2 )}$ in partial fractions.

1

Check: Numerator degree 3, denominator degree 2 (improper). Expand denominator:

x^{2} + x - 2

2

Divide

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

by

x^{2} + x - 2

: gives quotient

x

with remainder

2 x - 2

3

So:

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

4

Now partial fractions on the proper part:

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

5

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

. Let

x = 1

:

0 = 3 A

so

A = 0

6

Let

x = - 2

:

- 6 = - 3 B

so

B = 2

Answer

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

Practice

Question 1: Express

\frac{11 x - 1}{( x - 2 ) ( x + 1 ) ( x - 3 )}

in partial fractions.

Step 1: Write

\frac{11 x - 1}{( x - 2 ) ( x + 1 ) ( x - 3 )} = \frac{A}{x - 2} + \frac{B}{x + 1} + \frac{C}{x - 3}

Step 2: Multiply through:

11 x - 1 = A (x + 1) (x - 3) + B (x - 2) (x - 3) + C (x - 2) (x + 1)

Step 3: Let

x = 2

:

21 = A (3) (- 1) = - 3 A

so

A = - 7

Step 4: Let

x = - 1

:

- 12 = B (- 3) (- 4) = 12 B

so

B = - 1

Step 5: Let

x = 3

:

32 = C (1) (4) = 4 C

so

C = 8

Answer

\frac{11 x - 1}{( x - 2 ) ( x + 1 ) ( x - 3 )} = \frac{- 7}{x - 2} - \frac{1}{x + 1} + \frac{8}{x - 3}

Practice

Question 2: Express

\frac{5 x ^{2} - 2 x - 1}{( x - 1 ) ^{2} ( x + 2 )}

in partial fractions.

Step 1: Write

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

Step 2: Multiply:

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

Step 3: Let

x = 1

:

5 - 2 - 1 = 3 B

so

B = \frac{2}{3}

Step 4: Let

x = - 2

:

20 + 4 - 1 = 9 C

so

C = \frac{23}{9}

Step 5: Compare

x^{2}

coefficients:

5 = A + C = A + \frac{23}{9}

so

A = \frac{22}{9}

Answer

\frac{5 x ^{2} - 2 x - 1}{( x - 1 ) ^{2} ( x + 2 )} = \frac{22}{9 ( x - 1 )} + \frac{2}{3 ( x - 1 ) ^{2}} + \frac{23}{9 ( x + 2 )}

Practice

Question 3: Express

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

in partial fractions.

Step 1: Improper fraction (degree 3 vs degree 2). Divide first.

Step 2:

x^{3} - x^{2} + x + 4 \div (x^{2} - 4) = x - 1

remainder

5 x

Step 3: So

\frac{x ^{3} - x ^{2} + x + 4}{x ^{2} - 4} = x - 1 + \frac{5 x}{x ^{2} - 4} = x - 1 + \frac{5 x}{( x - 2 ) ( x + 2 )}

Step 4: Partial fractions:

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

Step 5:

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

. Let

x = 2

:

10 = 4 A

so

A = \frac{5}{2}

Step 6: Let

x = - 2

:

- 10 = - 4 B

so

B = \frac{5}{2}

Answer

\frac{x ^{3} - x ^{2} + x + 4}{x ^{2} - 4} = x - 1 + \frac{5}{2 ( x - 2 )} + \frac{5}{2 ( x + 2 )}

Summary

Choosing the Correct Form

Denominator FactorPartial Fraction Form

Linear:

(a x + b)

\frac{A}{a x + b}

Repeated linear:

(a x + b)^{2}

\frac{A}{a x + b} + \frac{B}{( a x + b ) ^{2}}

Improper fractionDivide first, then decompose remainder

Exam Checklist

Step 1: Is it proper? If not, divide first

Step 2: Factorise the denominator completely

Step 3: Write the correct partial fraction form

Step 4: Use substitution and/or comparing coefficients

Step 5: Check by adding your fractions back together