Partial fractions

IB Mathematics AI HL· Unit 1: Number and Algebra, Topic 9· 15 min read

1. Basics: Proper vs Improper Rational Functions★★☆☆☆HL only⏱ 5 min

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📘 Definition

Partial Fraction Decomposition

$\frac{P ( x )}{Q ( x )} = \sum \frac{A _{i}}{( a x + b ) ^{k}} + \sum \frac{B _{j} x + C _{j}}{( c x ^{2} + d x + e ) ^{m}}$

The process of breaking a complex rational function into a sum of simpler fractions that are easier to integrate or manipulate algebraically

Example:

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

Before starting decomposition, always check the degree of the numerator and denominator. For improper rational functions (degree of numerator ≥ degree of denominator), you must first perform polynomial long division to get a quotient polynomial plus a proper rational remainder, which you then decompose.

📐 Worked Example

Decompose $\frac{x ^{3} + 3 x ^{2} - 2 x + 1}{x ^{2} - 1}$ into partial fractions

1
Check degrees: numerator degree = 3, denominator degree = 2. Improper, so first do polynomial division.
2
Divide to get:
3
$x^3 + 3x^2 - 2x + 1 = (x^2 - 1)(x+3) + (-x + 4)$
4
Rewrite the original function as:
5
$x + 3 + \frac{-x + 4}{x^2 - 1}$
6
Factor the denominator of the proper remainder: $x^{2} - 1 = (x - 1) (x + 1)$ , two distinct linear factors.
7
Set up decomposition: $\frac{- x + 4}{( x - 1 ) ( x + 1 )} = \frac{A}{x - 1} + \frac{B}{x + 1}$ . Multiply both sides by $(x - 1) (x + 1)$ :
8
$-x + 4 = A(x+1) + B(x-1)$
9
Substitute $x = 1$ : $3 = 2 A \to A = \frac{3}{2}$ . Substitute $x = - 1$ : $5 = - 2 B \to B = - \frac{5}{2}$ .
10
Final decomposition:
11
$\frac{x^3 + 3x^2 - 2x + 1}{x^2 - 1} = x + 3 + \frac{3}{2(x-1)} - \frac{5}{2(x+1)}$

Exam tip:

Always check degrees first — forgetting to divide for improper functions is the most common exam mistake that costs easy marks.

2. Distinct and Repeated Linear Factors★★★☆☆HL only⏱ 6 min

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When the denominator factors into linear terms, different rules apply for distinct vs repeated factors:

Each distinct linear factor $(a x + b)$ gets one partial fraction with constant numerator $\frac{A}{a x + b}$
A repeated linear factor $(a x + b)^{n}$ gets $n$ partial fractions, one for each power from 1 to $n$

📘 Definition

Repeated Linear Factor

A linear factor of the denominator that is raised to a power greater than 1

Example:

For $\frac{2 x + 3}{( x - 1 ) ( x + 4 ) ^{2}}$ , the repeated factor $(x + 4)^{2}$ gives two partial fractions: $\frac{A}{x + 4} + \frac{B}{( x + 4 ) ^{2}}$

📐 Worked Example

Decompose $\frac{5 x ^{2} - 8 x + 3}{( x - 1 ) ( x - 2 ) ^{2}}$ into partial fractions

1
Check degrees: $2 < 3$ , so proper, no division needed.
2
Set up decomposition for one distinct and one repeated linear factor:
3
$\frac{5x^2 - 8x + 3}{(x-1)(x-2)^2} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{(x-2)^2}$
4
Multiply both sides by $(x - 1) (x - 2)^{2}$ :
5
$5x^2 - 8x + 3 = A(x-2)^2 + B(x-1)(x-2) + C(x-1)$
6
Substitute $x = 1$ : $0 = A (1) \to A = 0$ . Substitute $x = 2$ : $7 = C (1) \to C = 7$ .
7
Equate coefficients of $x^{2}$ : $5 = A + B \to B = 5$ .
8
Final decomposition:
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$\frac{5x^2 - 8x + 3}{(x-1)(x-2)^2} = \frac{5}{x-2} + \frac{7}{(x-2)^2}$

3. Irreducible Quadratic Factors★★★★☆HL only⏱ 7 min

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An irreducible quadratic factor is a quadratic $a x^{2} + b x + c$ that cannot be factored into linear terms with real coefficients, which occurs when its discriminant $b^{2} - 4 a c < 0$ . For these factors, you must use a linear numerator (not a constant) in the corresponding partial fraction.

📐 Worked Example

Decompose $\frac{4 x ^{2} + 13 x + 10}{( x + 1 ) ( x ^{2} + 2 x + 3 )}$ into partial fractions

1
Check: degree 2 < 3, proper. Check discriminant of quadratic: $2^{2} - 4 (1) (3) = - 8 < 0$ , so irreducible.
2
Set up decomposition with linear numerator for the quadratic factor:
3
$\frac{4x^2 + 13x + 10}{(x+1)(x^2 + 2x + 3)} = \frac{A}{x+1} + \frac{Bx + C}{x^2 + 2x + 3}$
4
Multiply through by the full denominator:
5
$4x^2 + 13x + 10 = A(x^2 + 2x + 3) + (Bx + C)(x+1)$
6
Substitute $x = - 1$ : $1 = 2 A \to A = \frac{1}{2}$ .
7
Expand RHS and equate coefficients: $x^{2} : 4 = A + B \to B = \frac{7}{2}$ , constant term: $10 = 3 A + C \to C = \frac{17}{2}$ .
8
Final decomposition:
9
$\frac{4x^2 + 13x + 10}{(x+1)(x^2 + 2x + 3)} = \frac{1}{2(x+1)} + \frac{7x + 17}{2(x^2 + 2x + 3)}$

4. Application: Integration of Rational Functions★★★☆☆HL only⏱ 5 min

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The primary use of partial fractions in IB AI HL is to simplify rational functions so they can be integrated using standard logarithmic rules for linear terms.

📐 Worked Example

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

1
Factor denominator: $x^{2} - x - 2 = (x - 2) (x + 1)$
2
Decompose to get: $\frac{3 x + 1}{( x - 2 ) ( x + 1 )} = \frac{7}{3 ( x - 2 )} + \frac{2}{3 ( x + 1 )}$
3
Rewrite the integral:
4
$\int \left( \frac{7}{3(x-2)} + \frac{2}{3(x+1)} \right) dx$
5
Integrate term by term using $\int \frac{1}{u} d u = ln ∣ u ∣ + C$ :
6
$\frac{7}{3} \ln|x-2| + \frac{2}{3} \ln|x+1| + C$

Exam tip:

Always include the absolute value inside the logarithm after integration — you will lose an accuracy mark if you omit it.

5. Common Pitfalls

Wrong move:

Forgetting polynomial division for improper rational functions

Why:

The partial fraction formula only works for proper rational functions, so you will get incorrect constant values

Correct move:

Always compare degrees first, divide first if numerator degree ≥ denominator degree

Wrong move:

Using a constant numerator for an irreducible quadratic factor

Why:

This creates an imbalance in the degrees of the equation, leading to wrong solutions

Correct move:

Always use a linear numerator $B x + C$ for irreducible quadratic factors

Wrong move:

Only adding one partial fraction for a repeated linear factor

Why:

You miss the lower power terms, leading to an incorrect decomposition

Correct move:

For $(a x + b)^{n}$ , add $n$ partial fractions, one for each power from 1 to $n$

Wrong move:

Omitting absolute value when integrating partial fractions

Why:

The natural logarithm is only defined for positive inputs, so omitting it is incorrect for negative $x$ values

Correct move:

Always write $ln ∣ a x + b ∣$ , not $ln (a x + b)$ , after integration

6. Quick Reference Cheatsheet

Denominator Factor Type	Corresponding Partial Fraction Term(s)
Distinct linear $(a x + b)$	$\frac{A}{a x + b}$
Repeated linear $(a x + b)^{n}$	$\frac{A _{1}}{a x + b} + \frac{A _{2}}{( a x + b ) ^{2}} + ... + \frac{A _{n}}{( a x + b ) ^{n}}$
Distinct irreducible quadratic $(a x^{2} + b x + c)$	$\frac{A x + B}{a x ^{2} + b x + c}$
Repeated irreducible quadratic $(a x^{2} + b x + c)^{n}$	$\frac{A _{1} x + B _{1}}{a x ^{2} + b x + c} + ... + \frac{A _{n} x + B _{n}}{( a x ^{2} + b x + c ) ^{n}}$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2025 · 1
Full partial fraction decomposition
2024 · 2
Partial fractions for definite integration
2023 · 1
Improper function decomposition

Going deeper

syllabusOfficial IB AI HL SyllabusUnit 1 learning outcomes reference

What's Next

Partial fractions are a foundational algebraic skill that you will use repeatedly when integrating rational functions, which appear in both Paper 1 and Paper 2 of IB AI HL exams. Mastering decomposition now simplifies all future integration work, and builds on your core polynomial algebra skills that apply across the entire syllabus. This topic connects directly to integration techniques, which are a major assessment component in IB AI HL, so solidifying your understanding here will pay off in later units.