What is too few terms for a repeated factor?

A factor (x - a)^2 needs both (A)/(x-a) and (B)/((x-a)^2). Using only one term loses a degree of freedom.

What is a constant numerator over an irreducible quadratic?

An irreducible quadratic factor needs a linear numerator Bx + C, not just a constant.

NESA 2024 HSC-style practice: Use partial fractions to find integral of (3x^2 + 2x + 1)/((x - 1)(x^2 + 1)) dx.

The denominator has a linear factor and an irreducible quadratic factor, so write (3x^2 + 2x + 1)/((x - 1)(x^2 + 1)) = A/(x - 1) + (Bx + C)/(x^2 + 1). Then 3x^2 + 2x + 1 = A(x^2 + 1) + (Bx + C)(x - 1). - At x = 1: 3 + 2 + 1 = 6 = 2A, so A = 3. - At x = 0: 1 = A - C = 3 - C, so C = 2. - Comparing x^2 coefficients: 3 = A + B = 3 + B, so B = 0. The integrand is 3/(x - 1) + 2/(x^2 + 1), so integral = 3 ln|x - 1| + 2 arctan(x) + C. Mark notes: 1 mark for the correct partial fraction form with a linear numerator over the quadratic, 1 mark for finding A = 3, B = 0, C = 2, 1 mark for the primitive 3 ln|x - 1| + 2 arctan(x) + C.

NESA 2021 HSC-style practice: Express (3x^2 - 5)/((x - 2)(x^2 + x + 1)) as a sum of partial fractions over the real numbers.

Since x^2 + x + 1 is irreducible over the reals (its discriminant 1 - 4 = -3 < 0), use a linear numerator over it: (3x^2 - 5)/((x - 2)(x^2 + x + 1)) = A/(x - 2) + (Bx + C)/(x^2 + x + 1). Multiply through: 3x^2 - 5 = A(x^2 + x + 1) + (Bx + C)(x - 2). - At x = 2: 12 - 5 = 7 = A(4 + 2 + 1) = 7A, so A = 1. - Comparing x^2 coefficients: 3 = A + B = 1 + B, so B = 2. - Comparing constants: -5 = A - 2C = 1 - 2C, so C = 3. (Check the x coefficient: A - 2B + C = 1 - 4 + 3 = 0, matching the 0 coefficient of x on the left.) So (3x^2 - 5)/((x - 2)(x^2 + x + 1)) = 1/(x - 2) + (2x + 3)/(x^2 + x + 1). Mark notes: 1 mark for the correct form with Bx + C over the irreducible quadratic, 1 mark for A = 1, 1 mark for B = 2 and C = 3.

NSWMaths Extension 2Syllabus dot point

How does decomposing a rational function into partial fractions turn an awkward integral into a sum of standard logarithm and arctangent integrals?

Decompose rational functions into partial fractions and use the decomposition to integrate, including linear, repeated and irreducible quadratic factors

A focused answer to the HSC Maths Extension 2 dot point on partial fractions. Decomposing proper rational functions over distinct linear, repeated and irreducible quadratic factors, then integrating to logs and arctangents, with verified worked examples.

Generated by Claude Opus 4.76 min answerUpdated 2026-05-29

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

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What this dot point is asking
When to use partial fractions
Distinct linear factors
Repeated linear factors
Irreducible quadratic factors

What this dot point is asking

NESA wants you to integrate rational functions (a polynomial divided by a polynomial) by first splitting them into simpler partial fractions. You must handle distinct linear factors, repeated linear factors, and irreducible quadratic factors, and integrate the resulting pieces into logarithms and arctangents.

When to use partial fractions

The method applies to proper rational functions, where the numerator has lower degree than the denominator. If the function is improper (numerator degree greater than or equal to denominator degree), first divide using polynomial long division to extract a polynomial plus a proper remainder, then decompose the remainder.

Distinct linear factors

If the denominator factorises into distinct linear factors, write one fraction per factor:

\frac{px + q}{(x - a)(x - b)} = \frac{A}{x - a} + \frac{B}{x - b}.

Multiply through by $(x - a)(x - b)$ and either equate coefficients or substitute convenient values of $x$ (the cover-up method): setting $x = a$ isolates $A$ , and $x = b$ isolates $B$ .

Each term integrates as $\int \dfrac{A}{x - a}\,dx = A\ln|x - a| + C$ .

Repeated linear factors

A repeated factor $(x - a)^2$ requires a term for each power up to the multiplicity:

\frac{\dots}{(x - a)^2} = \frac{A}{x - a} + \frac{B}{(x - a)^2}.

The first term integrates to a logarithm, the second to $-\dfrac{B}{x - a}$ via the power rule.

Irreducible quadratic factors

A quadratic $x^2 + px + q$ with no real roots (negative discriminant) gets a linear numerator:

\frac{\dots}{x^2 + px + q} = \frac{Bx + C}{x^2 + px + q}.

Such a term integrates by splitting into a part proportional to the derivative of the denominator (giving a logarithm) plus a constant part, which after completing the square gives an arctangent.

Worked example

Evaluate $\displaystyle\int \frac{3x + 1}{(x - 1)(x + 2)}\,dx$

Decompose: $\dfrac{3x + 1}{(x - 1)(x + 2)} = \dfrac{A}{x - 1} + \dfrac{B}{x + 2}$ . Multiply through:

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

Set $x = 1$ : $4 = A(3)$ , so $A = \dfrac{4}{3}$ . Set $x = -2$ : $-5 = B(-3)$ , so $B = \dfrac{5}{3}$ .

Therefore

\int \frac{3x + 1}{(x - 1)(x + 2)}\,dx = \frac{4}{3}\ln|x - 1| + \frac{5}{3}\ln|x + 2| + C.

Check the decomposition at $x = 0$ : left side $\dfrac{1}{(-1)(2)} = -\dfrac{1}{2}$ ; right side $\dfrac{4/3}{-1} + \dfrac{5/3}{2} = -\dfrac{4}{3} + \dfrac{5}{6} = -\dfrac{8}{6} + \dfrac{5}{6} = -\dfrac{3}{6} = -\dfrac{1}{2}$ . They match.

Exam-style practice questions

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

2023 HSC4 marksFind integral from 0 to 2 of (5x - 3)/((x + 1)(x - 3)) dx.

Show worked answer →

Decompose into partial fractions over the distinct linear factors: (5x - 3)/((x + 1)(x - 3)) = A/(x + 1) + B/(x - 3).

Then 5x - 3 = A(x - 3) + B(x + 1).

At x = 3: 15 - 3 = 12 = 4B, so B = 3.
At x = -1: -5 - 3 = -8 = -4A, so A = 2.

So the integrand is 2/(x + 1) + 3/(x - 3), and

integral from 0 to 2 = [2 ln|x + 1| + 3 ln|x - 3|] from 0 to 2.

At x = 2: 2 ln 3 + 3 ln 1 = 2 ln 3. At x = 0: 2 ln 1 + 3 ln 3 = 3 ln 3.

Value = 2 ln 3 - 3 ln 3 = -ln 3 (equivalently ln(1/3)).

Mark notes: 1 mark for the partial fraction setup, 1 mark for A = 2 and B = 3, 1 mark for the antiderivative with logs, 1 mark for evaluating to -ln 3 (note |x - 3| keeps the logs real on [0, 2]).

2024 HSC3 marksUse partial fractions to find integral of (3x^2 + 2x + 1)/((x - 1)(x^2 + 1)) dx.

Show worked answer →

The denominator has a linear factor and an irreducible quadratic factor, so write

(3x^2 + 2x + 1)/((x - 1)(x^2 + 1)) = A/(x - 1) + (Bx + C)/(x^2 + 1).

Then 3x^2 + 2x + 1 = A(x^2 + 1) + (Bx + C)(x - 1).

At x = 1: 3 + 2 + 1 = 6 = 2A, so A = 3.
At x = 0: 1 = A - C = 3 - C, so C = 2.
Comparing x^2 coefficients: 3 = A + B = 3 + B, so B = 0.

The integrand is 3/(x - 1) + 2/(x^2 + 1), so

integral = 3 ln|x - 1| + 2 arctan(x) + C.

Mark notes: 1 mark for the correct partial fraction form with a linear numerator over the quadratic, 1 mark for finding A = 3, B = 0, C = 2, 1 mark for the primitive 3 ln|x - 1| + 2 arctan(x) + C.

2021 HSC3 marksExpress (3x^2 - 5)/((x - 2)(x^2 + x + 1)) as a sum of partial fractions over the real numbers.

Show worked answer →

Since x^2 + x + 1 is irreducible over the reals (its discriminant 1 - 4 = -3 < 0), use a linear numerator over it:

(3x^2 - 5)/((x - 2)(x^2 + x + 1)) = A/(x - 2) + (Bx + C)/(x^2 + x + 1).

Multiply through: 3x^2 - 5 = A(x^2 + x + 1) + (Bx + C)(x - 2).

At x = 2: 12 - 5 = 7 = A(4 + 2 + 1) = 7A, so A = 1.
Comparing x^2 coefficients: 3 = A + B = 1 + B, so B = 2.
Comparing constants: -5 = A - 2C = 1 - 2C, so C = 3.

(Check the x coefficient: A - 2B + C = 1 - 4 + 3 = 0, matching the 0 coefficient of x on the left.)

So (3x^2 - 5)/((x - 2)(x^2 + x + 1)) = 1/(x - 2) + (2x + 3)/(x^2 + x + 1).

Mark notes: 1 mark for the correct form with Bx + C over the irreducible quadratic, 1 mark for A = 1, 1 mark for B = 2 and C = 3.