Partial fractions (SACE Stage 2 Specialist Mathematics)

Syllabus dot point

“Decompose rational functions into partial fractions and use the decomposition to integrate them.”

8 min readUpdated 2 Jun 2026

Quick answer

Partial fractions rewrite a rational function as a sum of simpler fractions, one per factor of the denominator. For distinct linear factors, $\dfrac{P(x)}{(x-a)(x-b)} = \dfrac{A}{x-a} + \dfrac{B}{x-b}$ ; find $A$ and $B$ by substituting convenient values or equating coefficients. Each simple fraction integrates to a logarithm, so the technique turns hard rational integrals into easy ones.

Jump to a section

What this dot point is asking
When and how to decompose
Finding the constants
Integrating the pieces
Equating coefficients
Improper rational functions
Connecting to differential equations
Exam-style questions

What this dot point is asking

You need to decompose rational functions into partial fractions and use the decomposition to integrate them.

When and how to decompose

Partial fractions apply to a proper rational function (numerator degree less than denominator degree). If the function is improper, do polynomial division first and decompose the remaining proper part.

Form of the decomposition

A distinct linear factor $(x - a)$ contributes a term $\dfrac{A}{x - a}$ .
A repeated linear factor $(x - a)^2$ contributes $\dfrac{A}{x - a} + \dfrac{B}{(x - a)^2}$ .
An irreducible quadratic factor $(x^2 + px + q)$ contributes $\dfrac{Ax + B}{x^2 + px + q}$ .
Set up one term per factor, then solve for the unknown constants.

Finding the constants

Two reliable methods: substitute strategic $x$ -values (the cover-up idea), or expand and equate coefficients of like powers.

Decomposing into partial fractions

Express $\dfrac{5x - 4}{(x - 2)(x + 1)}$ in partial fractions

Step 1: Set up the partial fraction form

Because the denominator has two distinct linear factors, one term per factor is assigned an unknown numerator. The idea is that any proper rational function over distinct linear factors can be written as a sum of simpler fractions of exactly this shape.

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

Step 2: Clear the denominators

Multiplying both sides by $(x-2)(x+1)$ eliminates the fractions and leaves a polynomial identity that holds for all values of $x$ .

5x - 4 = A(x + 1) + B(x - 2).

Step 3: Substitute $x = 2$ to find $A$

Substituting a root of one factor makes the term containing the other unknown vanish, isolating $A$ . Here $x = 2$ is a root of $(x - 2)$ , so the $B$ term disappears.

$5(2) - 4 = A(3)$ , so $6 = 3A$ , giving $A = 2$ .

Step 4: Substitute $x = -1$ to find $B$ , then write the result

Substituting $x = -1$ , a root of $(x + 1)$ , makes the $A$ term vanish and isolates $B$ .

$5(-1) - 4 = B(-3)$ , so $-9 = -3B$ , giving $B = 3$ .

Therefore

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

Integrating the pieces

Each $\dfrac{A}{x - a}$ integrates to $A\ln|x - a|$ , using $\int \dfrac{1}{x - a}\,dx = \ln|x - a| + C$ . So once decomposed, the integral is a sum of logarithms.

Integrating using partial fractions

Find $\displaystyle\int \frac{5x - 4}{(x-2)(x+1)}\,dx$

Step 1: Use the decomposition from the previous example to rewrite the integral

The hard part of integration has already been done: partial fractions replaced the rational function with simpler pieces. Each piece is of the form $\dfrac{A}{x - a}$ , which has a known antiderivative.

\int \frac{5x - 4}{(x-2)(x+1)}\,dx = \int \left(\frac{2}{x - 2} + \frac{3}{x + 1}\right)dx.

Step 2: Integrate each term using the standard logarithm rule

The rule $\int \dfrac{1}{x - a}\,dx = \ln|x - a| + C$ applies directly to each term. Integrating term by term and collecting the constant gives the final answer.

= 2\ln|x - 2| + 3\ln|x + 1| + C.

Equating coefficients

The substitution (cover-up) method is fastest for distinct linear factors, but equating coefficients is the general approach and is essential when a repeated or quadratic factor is present. After clearing denominators, expand the right side fully and match the coefficients of each power of $x$ on both sides. This produces a small system of simultaneous equations in the unknown constants, which you solve together. For example, $3x + 5 = A(x+1) + B(x-2)$ gives, on matching the $x$ coefficients and the constants, $A + B = 3$ and $A - 2B = 5$ , solved to find $A$ and $B$ . Both methods give the same answer, so use whichever the factor structure makes cleaner.

Improper rational functions

If the numerator degree is at least the denominator degree, divide first. For instance $\dfrac{x^2}{x^2 - 1}$ is improper; dividing gives $1 + \dfrac{1}{x^2 - 1}$ , and only the proper remainder $\dfrac{1}{(x-1)(x+1)}$ is decomposed.

Always factor the denominator first

Partial fractions need the denominator fully factored into linear and irreducible quadratic factors. Spotting a difference of squares such as $x^2 - 1 = (x-1)(x+1)$ is often the first step.

Connecting to differential equations

Partial fractions are the workhorse behind solving logistic differential equations in Topic 6. An equation such as $\dfrac{dP}{dt} = kP(M - P)$ separates to $\dfrac{1}{P(M-P)}\,dP = k\,dt$ , and the left side is integrated only after decomposing $\dfrac{1}{P(M-P)}$ into $\dfrac{1}{M}\left(\dfrac{1}{P} + \dfrac{1}{M-P}\right)$ . Each piece then integrates to a logarithm, producing the $\ln\dfrac{P}{M-P}$ form that rearranges into the S-shaped logistic solution. Mastery of partial fractions here pays off directly when you model limited growth.

Common errors

Decomposing an improper fraction directly: If the numerator degree is not smaller, divide first; otherwise the decomposition cannot match.
Wrong form for repeated factors: A factor $(x - a)^2$ needs both $\dfrac{A}{x-a}$ and $\dfrac{B}{(x-a)^2}$ ; using only one term loses solutions.
Forgetting the absolute value in the log: $\int \dfrac{1}{x-a}\,dx = \ln|x - a| + C$ . Dropping the modulus is a marked error.
Arithmetic slips finding constants: After substituting, double-check by recombining the partial fractions; they should return the original numerator.

Exam-style questions

Questions in the style of SACE Board exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

SACE 2022-style2 marks

Calculator-free. Show that

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

Show worked answer →

Combine the left side over the common denominator $(x-2)(x+3)$ :

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

The numerator simplifies: $(x+3) - (x-2) = 5$ . So the left side equals $\dfrac{5}{(x-2)(x+3)}$ , as required. This is the partial-fraction decomposition read in reverse. [2 marks]

SACE 2023-style2 marks

Calculator-free. Express

\dfrac{1}{(x-8)(x-4)}

in partial fractions.

Show worked answer →

Set $\dfrac{1}{(x-8)(x-4)} = \dfrac{A}{x-8} + \dfrac{B}{x-4}$ . Multiply through: $1 = A(x-4) + B(x-8)$ . [1 mark]

Let $x = 8$ : $1 = 4A$ , so $A = \tfrac{1}{4}$ . Let $x = 4$ : $1 = -4B$ , so $B = -\tfrac{1}{4}$ .

Therefore $\dfrac{1}{(x-8)(x-4)} = \dfrac{1}{4}\left(\dfrac{1}{x-8} - \dfrac{1}{x-4}\right)$ . [1 mark]

SACE 2021-style4 marks

Calculator-free. Express

\dfrac{5x-4}{(x-2)(x+1)}

in partial fractions and hence find

\displaystyle\int \dfrac{5x-4}{(x-2)(x+1)}\,dx

Show worked answer →

Set $\dfrac{5x-4}{(x-2)(x+1)} = \dfrac{A}{x-2} + \dfrac{B}{x+1}$ , so $5x - 4 = A(x+1) + B(x-2)$ .

Let $x = 2$ : $6 = 3A$ , so $A = 2$ . Let $x = -1$ : $-9 = -3B$ , so $B = 3$ . [2 marks]

Integrate term by term, using $\int \dfrac{1}{x-a}\,dx = \ln|x-a|$ :

\int \left(\frac{2}{x-2} + \frac{3}{x+1}\right)dx = 2\ln|x-2| + 3\ln|x+1| + C. \;[\text{2 marks}]

Practise this

Sources & how we know this

Stage 2 Specialist Mathematics subject outline (2025 onwards) — SACE Board
Stage 2 Specialist Mathematics: external assessment — SACE Board

What this dot point is asking

When and how to decompose

Finding the constants

Express 5x−4(x−2)(x+1)\dfrac{5x - 4}{(x - 2)(x + 1)}(x−2)(x+1)5x−4​ in partial fractions

Step 1: Set up the partial fraction form

Step 2: Clear the denominators

Step 3: Substitute x=2x = 2x=2 to find AAA

Step 4: Substitute x=−1x = -1x=−1 to find BBB, then write the result

Integrating the pieces

Find ∫5x−4(x−2)(x+1) dx\displaystyle\int \frac{5x - 4}{(x-2)(x+1)}\,dx∫(x−2)(x+1)5x−4​dx

Step 1: Use the decomposition from the previous example to rewrite the integral

Step 2: Integrate each term using the standard logarithm rule

Equating coefficients

Improper rational functions

Connecting to differential equations

Exam-style questions

Practise this

Sources & how we know this

Express $\dfrac{5x - 4}{(x - 2)(x + 1)}$ in partial fractions

Step 3: Substitute $x = 2$ to find $A$

Step 4: Substitute $x = -1$ to find $B$ , then write the result

Find $\displaystyle\int \frac{5x - 4}{(x-2)(x+1)}\,dx$