What are wrong form for repeated factors?

A factor (x - a)^2 needs both Ax-a and B(x-a)^2; using only one term loses solutions.

What are arithmetic slips finding constants?

After substituting, double-check by recombining the partial fractions; they should return the original numerator.

SACE Board SACE 2022-style practice: Calculator-free. Show that 1x-2 - 1x+3 = 5(x-2)(x+3).

Combine the left side over the common denominator (x-2)(x+3): $(1)/(x-2) - (1)/(x+3) = ((x+3) - (x-2))/((x-2)(x+3)).$ The numerator simplifies: (x+3) - (x-2) = 5. So the left side equals 5(x-2)(x+3), as required. This is the partial-fraction decomposition read in reverse. [2 marks]

SACE Board SACE 2023-style practice: Calculator-free. Express 1(x-8)(x-4) in partial fractions.

Set 1(x-8)(x-4) = Ax-8 + Bx-4. Multiply through: 1 = A(x-4) + B(x-8). [1 mark] Let x = 8: 1 = 4A, so A = 14. Let x = 4: 1 = -4B, so B = -14. Therefore 1(x-8)(x-4) = 14(1x-8 - 1x-4). [1 mark]

§-Syllabus dot point

SASpecialist MathematicsSyllabus dot point

How does splitting a rational function into simpler fractions make it integrable?

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

Decomposing a proper rational function into partial fractions over linear factors, finding the unknown constants, and integrating the pieces to logarithms.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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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

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.

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.

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.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SACE 20222 marksCalculator-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 20232 marksCalculator-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 20214 marksCalculator-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}]