What is too few terms for a repeated factor?

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

What is a constant numerator over an irreducible quadratic?

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

NESA HSC 2024-style practice: Use partial fractions to find (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: 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 int = 3ln|x - 1| + 2 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 3ln|x - 1| + 2 x + C.

NESA HSC 2021-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: 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.

§-Syllabus dot point

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, the cover-up method, and integrating to logs and arctangents, with verified worked examples.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-20

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
The cover-up method
Finding the unknown constants
Integrating each piece
How exam questions ask about partial fractions

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 recognise the three cases (distinct linear factors, repeated linear factors, and irreducible quadratic factors), write the correct decomposition form for each, find the unknown constants, and integrate the resulting pieces into logarithms and arctangents. The skill is examined both as a standalone "express as partial fractions" question and as the engine inside a larger integral.

When to use partial fractions

The method applies to proper rational functions, where the numerator has lower degree than the denominator. That degree condition is not a technicality, it is what guarantees a decomposition exists with constant (and linear) numerators. If the function is improper (numerator degree greater than or equal to denominator degree), you must first divide using polynomial long division to extract a polynomial part plus a proper remainder, then decompose only the remainder. Skipping the division on an improper fraction produces a system of equations with no solution, which is a sure sign you have started in the wrong place.

The deeper reason the method is worth the effort: a single complicated rational function has no obvious antiderivative, but each partial fraction is a textbook standard form. So decomposition converts a hard problem into a sum of easy ones, and you can predict the shape of the answer (logarithms from linear factors, arctangents from irreducible quadratics) before doing any integrating.

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 $(x - a)(x - b)$ to clear denominators, then either equate coefficients or substitute convenient values of $x$ . Each term integrates as $\int \dfrac{A}{x - a}\,dx = A\ln|x - a| + C$ .

Repeated linear factors

A repeated factor $(x - a)^2$ needs a separate term for every power up to the multiplicity, not just one term:

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

The reason both terms are needed is a counting argument: a proper fraction over $(x - a)^2$ has two degrees of freedom in its numerator, so the decomposition must also carry two unknowns. Using only $\dfrac{B}{(x-a)^2}$ throws one away and the system fails. A cube $(x - a)^3$ would need three terms, up to $\dfrac{C}{(x - a)^3}$ . The first term integrates to a logarithm; the higher terms integrate by the power rule, for example $\int \dfrac{B}{(x - a)^2}\,dx = -\dfrac{B}{x - a}$ .

Irreducible quadratic factors

A quadratic $x^2 + px + q$ with no real roots (negative discriminant, $p^2 - 4q < 0$ ) cannot be split into real linear factors, so it gets a full linear numerator:

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

A constant numerator is not enough here, for the same degrees-of-freedom reason: a proper fraction over a quadratic has a numerator of degree up to one, so you need both $B$ and $C$ . Such a term integrates in two pieces: split it into a part proportional to the derivative $2x + p$ of the denominator (which integrates to a logarithm) plus a leftover constant, which after completing the square integrates to an arctangent.

The cover-up method

For distinct linear factors there is a shortcut that avoids solving a system. To find the constant over $(x - a)$ , multiply the whole identity by $(x - a)$ and then set $x = a$ : every other term still has a factor of $(x - a)$ in it and vanishes, leaving the constant alone. In practice you "cover up" the factor $(x - a)$ in the original denominator and evaluate what remains at $x = a$ . For $\dfrac{5x - 3}{(x + 1)(x - 3)}$ , the constant over $(x + 1)$ is found by covering $(x + 1)$ and putting $x = -1$ into $\dfrac{5x - 3}{x - 3} = \dfrac{-8}{-4} = 2$ . The cover-up method is fast and reliable for simple linear factors, but it only isolates constants over single linear factors; for repeated or irreducible-quadratic blocks you still need to equate coefficients for the remaining unknowns.

Finding the unknown constants

Two methods recover the constants $A$ , $B$ , $C$ after you clear the denominators, and good answers mix them. The cover-up (substitution) method substitutes the value of $x$ that makes one factor zero, instantly isolating the constant over that factor; it is fastest for distinct linear factors. The equate-coefficients method expands the right side and matches the coefficient of each power of $x$ on both sides, producing a small linear system; it is the reliable route when irreducible quadratics or repeated factors are involved, since not every constant can be isolated by a single substitution. The efficient workflow is to cover up the linear factors first, then equate one or two coefficients (often the highest power and the constant term) to pin down the rest. Always verify with a check value of $x$ , as in the worked examples below, before integrating.

Integrating each piece

Once decomposed, each fragment maps to a standard integral, and recognising the target shape in advance is a useful sanity check on your decomposition.

$\displaystyle\int \frac{A}{x - a}\,dx = A\ln|x - a| + C$ (a logarithm; keep the absolute value).
$\displaystyle\int \frac{B}{(x - a)^2}\,dx = -\frac{B}{x - a} + C$ (power rule, no logarithm).
$\displaystyle\int \frac{Bx + C}{x^2 + px + q}\,dx$ : write the numerator as $\tfrac{B}{2}(2x + p)$ plus a constant. The first part gives $\tfrac{B}{2}\ln(x^2 + px + q)$ (no absolute value, since an irreducible quadratic is always positive); the leftover constant over the completed square gives an arctangent.

So the final answer to a partial-fractions integral is always a combination of logarithms and arctangents (plus possibly a polynomial, if the original fraction was improper).

How exam questions ask about partial fractions

"Express $\dfrac{\ldots}{\ldots}$ as a sum of partial fractions." Decomposition only, no integration. Choose the correct form for each factor and find all constants; finish with the constants substituted in.
"Use partial fractions to find $\int \ldots$ ." Decompose first, then integrate term by term to logs and arctangents, with $+C$ .
"Hence find $\int_a^b \ldots$ " (definite). Decompose, integrate, then substitute the limits; the modulus inside the logs keeps everything real over the interval.
A denominator with an irreducible quadratic ( $x^2 + 1$ , $x^2 + x + 1$ ). Use a linear numerator $Bx + C$ ; expect an arctangent in the answer.
A repeated factor $(x - a)^2$ or $(x + a)^2$ . Include both $\dfrac{A}{x - a}$ and $\dfrac{B}{(x - a)^2}$ ; the second integrates without a log.
An improper-looking integrand (numerator degree $\ge$ denominator degree). Do polynomial long division first, then decompose the remainder.

Evaluate

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

(distinct linear factors)

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

The integrand is a proper rational function with two distinct linear factors in the denominator. We split it into partial fractions, find the constants using the cover-up method, then integrate each piece as a logarithm.

Set up the partial fraction decomposition: Each distinct linear factor in the denominator contributes one fraction with a constant numerator. Write:

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

Multiply both sides through by $(x - 1)(x + 2)$ to clear the denominators:

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

Find the constants using the cover-up method: Substitute the value of $x$ that makes each factor zero. Setting $x = 1$ kills the $B$ term: $4 = A(3)$ , so $A = \dfrac{4}{3}$ . Setting $x = -2$ kills the $A$ term: $-5 = B(-3)$ , so $B = \dfrac{5}{3}$ .

Integrate each partial fraction: Each term of the form $\dfrac{C}{x - a}$ integrates to $C\ln|x - a|$ . Combining the two pieces:

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

Verify the decomposition: Check at $x = 0$ : the original integrand gives $\dfrac{1}{(-1)(2)} = -\dfrac{1}{2}$ ; the decomposition gives $\dfrac{4/3}{-1} + \dfrac{5/3}{2} = -\dfrac{4}{3} + \dfrac{5}{6} = -\dfrac{8}{6} + \dfrac{5}{6} = -\dfrac{1}{2}$ . The values match, confirming the constants are correct.

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

Evaluate

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

(a repeated linear factor)

Evaluate $\displaystyle\int \frac{x + 3}{(x - 1)^2}\,dx$ with a repeated linear factor

A repeated factor needs one term per power, so the decomposition carries two unknowns even though there is only one distinct factor.

Write the form for a repeated factor: The factor $(x - 1)^2$ contributes a term for each power up to the multiplicity:

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

Multiply through by $(x - 1)^2$ :

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

Find the constants: Cover-up still gives one of them: setting $x = 1$ kills the $A$ term, so $B = 1 + 3 = 4$ . For $A$ , compare the coefficient of $x$ on each side: the left has $1$ , the right has $A$ , so $A = 1$ . (Equivalently, substitute any other value of $x$ .)

Integrate, noting the power-rule term: The first term gives a logarithm; the second integrates by the power rule, not to a log:

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

Final answer: $\displaystyle\int \frac{x + 3}{(x - 1)^2}\,dx = \ln|x - 1| - \frac{4}{x - 1} + C$ .

Evaluate

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

(an irreducible quadratic)

Evaluate $\displaystyle\int \frac{3x^2 + 2x + 1}{(x - 1)(x^2 + 1)}\,dx$ with an irreducible quadratic

The denominator mixes a linear factor with an irreducible quadratic, so the answer will contain both a logarithm and an arctangent.

Write the form with a linear numerator over the quadratic: Since $x^2 + 1$ has discriminant $0 - 4 = -4 < 0$ it is irreducible, so it gets a $Bx + C$ numerator:

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

Multiply through by $(x - 1)(x^2 + 1)$ :

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

Find the constants by mixing cover-up and coefficients: Cover-up the linear factor: at $x = 1$ , $\;3 + 2 + 1 = A(2)$ , so $A = 3$ . Compare $x^2$ coefficients: $3 = A + B = 3 + B$ , so $B = 0$ . Substitute $x = 0$ for the constant: $1 = A - C = 3 - C$ , so $C = 2$ .

Integrate each piece to a log and an arctangent: With $B = 0$ the quadratic piece is just $\dfrac{2}{x^2 + 1}$ , a standard arctangent form:

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

Final answer: $\displaystyle\int \frac{3x^2 + 2x + 1}{(x - 1)(x^2 + 1)}\,dx = 3\ln|x - 1| + 2\arctan x + C$ .

Edge case: an improper fraction,

\displaystyle\int \frac{x^2}{x^2 - 1}\,dx

Integrate the improper fraction $\displaystyle\int \frac{x^2}{x^2 - 1}\,dx$

The numerator degree equals the denominator degree, so this is improper and must be divided before any decomposition.

Divide first to extract the polynomial part: Because the degrees are equal, write $x^2 = (x^2 - 1) + 1$ , so

\frac{x^2}{x^2 - 1} = 1 + \frac{1}{x^2 - 1}.

The constant $1$ is the polynomial part; only the proper remainder $\dfrac{1}{x^2 - 1}$ is decomposed.

Decompose the proper remainder: Factor $x^2 - 1 = (x - 1)(x + 1)$ and write $\dfrac{1}{(x-1)(x+1)} = \dfrac{A}{x - 1} + \dfrac{B}{x + 1}$ . Cover-up gives $A = \dfrac{1}{2}$ (at $x = 1$ ) and $B = -\dfrac{1}{2}$ (at $x = -1$ ).

Integrate the polynomial part and the fractions:

\int \frac{x^2}{x^2 - 1}\,dx = \int 1\,dx + \int \frac{1/2}{x - 1}\,dx - \int \frac{1/2}{x + 1}\,dx = x + \tfrac{1}{2}\ln|x - 1| - \tfrac{1}{2}\ln|x + 1| + C.

This can be tidied with log laws to $x + \tfrac{1}{2}\ln\left|\dfrac{x - 1}{x + 1}\right| + C$ .

Final answer: $\displaystyle\int \frac{x^2}{x^2 - 1}\,dx = x + \tfrac{1}{2}\ln\left|\frac{x - 1}{x + 1}\right| + C$ .

Common traps

Decomposing an improper fraction directly: If the numerator degree is at least the denominator degree, divide first. Skipping this gives an inconsistent system with no solution.
Too few terms for a repeated factor: A factor $(x - a)^2$ needs both $\dfrac{A}{x-a}$ and $\dfrac{B}{(x-a)^2}$ . Using only one term loses a degree of freedom and the system fails.
A constant numerator over an irreducible quadratic: An irreducible quadratic factor needs a linear numerator $Bx + C$ , not just a constant.
Treating the repeated term as a logarithm: $\dfrac{B}{(x - a)^2}$ integrates by the power rule to $-\dfrac{B}{x - a}$ , not to a log.
Dropping the absolute value in the logarithm: Integrating $\dfrac{1}{x - a}$ gives $\ln|x - a|$ ; the modulus matters for negative arguments. (An irreducible quadratic is always positive, so its log needs no modulus.)

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.

HSC 20234 marksFind

\displaystyle\int_0^2 \frac{5x - 3}{(x + 1)(x - 3)} \, dx

Show worked answer →

Decompose into partial fractions over the distinct linear factors: $\frac{5x - 3}{(x + 1)(x - 3)} = \frac{A}{x + 1} + \frac{B}{x - 3}$ .

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

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

So the integrand is $\frac{2}{x + 1} + \frac{3}{x - 3}$ , and

$\int_0^2 = \left[2\ln|x + 1| + 3\ln|x - 3|\right]_0^2$ .

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 $\ln\frac{1}{3}$ ).

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 $[0, 2]$ ).

HSC 20243 marksUse partial fractions to find

\displaystyle\int \frac{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

$\frac{3x^2 + 2x + 1}{(x - 1)(x^2 + 1)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + 1}$ .

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

At $x = 1$ : $6 = 2A$ , 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 $\frac{3}{x - 1} + \frac{2}{x^2 + 1}$ , so

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

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 $3\ln|x - 1| + 2\arctan x + C$ .

HSC 20213 marksExpress

\dfrac{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:

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

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

At $x = 2$ : $7 = A(4 + 2 + 1) = 7A$ , 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 $\frac{3x^2 - 5}{(x - 2)(x^2 + x + 1)} = \frac{1}{x - 2} + \frac{2x + 3}{x^2 + x + 1}$ .

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 $C = 3$ .