Algebraic Fractions and Partial Fractions

This lesson covers simplifying algebraic fractions, performing arithmetic with them, and decomposing rational expressions into partial fractions — a technique essential for integration at A-Level.

Spec Mapping — AQA 7357 Section B Algebra and functions. This lesson covers the rational-expression simplification and partial-fraction decomposition content of Section B, the Year 2 (A2) extension of Pure 1's algebra strand. Refer to the official AQA specification document for exact wording.

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Simplifying Algebraic Fractions

To simplify an algebraic fraction, factorise the numerator and denominator, then cancel common factors.

Specimen Question (Modelled on the AQA Paper Format)

Note: this question is constructed to model AQA Paper 1/2/3 style; it is not a reproduction of any published past paper.

Worked Example 1

Simplify (x² − 9)/(x² + 5x + 6).

Numerator: x² − 9 = (x − 3)(x + 3)
Denominator: x² + 5x + 6 = (x + 2)(x + 3)

Cancel (x + 3):

= (x − 3)/(x + 2), provided x ≠ −3

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Adding and Subtracting Algebraic Fractions

Find a common denominator, combine, and simplify.

Worked Example 2

Express 3/(x + 1) − 2/(x − 3) as a single fraction.

Common denominator: (x + 1)(x − 3)

= [3(x − 3) − 2(x + 1)] / [(x + 1)(x − 3)]
= [3x − 9 − 2x − 2] / [(x + 1)(x − 3)]
= (x − 11) / [(x + 1)(x − 3)]

Answer: (x − 11) / [(x + 1)(x − 3)]

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Multiplying and Dividing Algebraic Fractions

Multiply: multiply numerators together and denominators together, then simplify.

Divide: multiply by the reciprocal.

Worked Example 3

Simplify [(x² − 4)/(x + 5)] × [(x + 5)/(x + 2)].

= [(x − 2)(x + 2)/(x + 5)] × [(x + 5)/(x + 2)]

Cancel (x + 5) and (x + 2):

= x − 2

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Improper Algebraic Fractions

An algebraic fraction is improper when the degree of the numerator is greater than or equal to the degree of the denominator. Use polynomial long division to express it as a polynomial plus a proper fraction.

Worked Example 4

Express (x³ + 2x² − x + 3)/(x + 1) in the form ax² + bx + c + d/(x + 1).

Perform polynomial division:

x³ + 2x² − x + 3 = (x + 1)(x² + x − 2) + 5

So: (x³ + 2x² − x + 3)/(x + 1) = x² + x − 2 + 5/(x + 1)

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Partial Fractions

Partial fractions decompose a proper fraction into simpler fractions. This is essential for integration and series work.

Type 1: Linear Factors

f(x)/[(x − a)(x − b)] = A/(x − a) + B/(x − b)

Worked Example 5

Express (5x + 3)/[(x + 1)(x − 2)] in partial fractions.

Let (5x + 3)/[(x + 1)(x − 2)] = A/(x + 1) + B/(x − 2)

Multiply through: 5x + 3 = A(x − 2) + B(x + 1)

Set x = 2: 13 = 3B → B = 13/3
Set x = −1: −2 = −3A → A = 2/3

= 2/[3(x + 1)] + 13/[3(x − 2)]

Type 2: Repeated Linear Factor

f(x)/[(x − a)²(x − b)] = A/(x − a) + B/(x − a)² + C/(x − b)

Worked Example 6

Express (3x² + x − 2)/[(x − 1)²(x + 2)] in partial fractions.

Let = A/(x − 1) + B/(x − 1)² + C/(x + 2)

Multiply: 3x² + x − 2 = A(x − 1)(x + 2) + B(x + 2) + C(x − 1)²

Set x = 1: 3 + 1 − 2 = 3B → B = 2/3
Set x = −2: 12 − 2 − 2 = 9C → C = 8/9
Compare x² coefficients: 3 = A + C → A = 3 − 8/9 = 19/9

= 19/[9(x − 1)] + 2/[3(x − 1)²] + 8/[9(x + 2)]

Type 3: Irreducible Quadratic Factor

f(x)/[(x − a)(x² + bx + c)] = A/(x − a) + (Bx + C)/(x² + bx + c)

Worked Example 7

Express (4x² + 1)/[(x − 1)(x² + 1)] in partial fractions.

Let = A/(x − 1) + (Bx + C)/(x² + 1)

Multiply: 4x² + 1 = A(x² + 1) + (Bx + C)(x − 1)

Set x = 1: 5 = 2A → A = 5/2
Compare x² coefficients: 4 = A + B → B = 4 − 5/2 = 3/2
Compare constants: 1 = A − C → C = 5/2 − 1 = 3/2

= 5/[2(x − 1)] + (3x + 3)/[2(x² + 1)]

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Synoptic Links

This topic connects to:

• Polynomial factorisation and the factor theorem (aqa-alevel-maths-pure-1 / algebra-and-functions) — required to factorise numerators and denominators before any cancellation.
• Integration using partial fractions (aqa-alevel-maths-pure-2 / integration-techniques) — partial-fraction decomposition is the gateway for integrating rational functions to logarithms.
• Binomial expansion for any rational index (aqa-alevel-maths-advanced-algebra / binomial-expansion) — splitting a fraction into partial pieces lets each piece be expanded separately.

Exam Tips

• Always factorise fully before simplifying or decomposing.
• For partial fractions, the cover-up method (substituting roots) is quick for linear factors.
• Check your answer by recombining the partial fractions.
• Identify the type of denominator first: distinct linear, repeated linear, or irreducible quadratic.
• Improper fractions must be divided out before applying partial fractions.

A-Level Deep Dive: Algebraic Fractions and Partial Fractions

Spec mapping

AQA 7357 Pure section B (Year 2) — Algebra and functions covers decompose rational functions into partial fractions (denominators not more complicated than squared linear terms and with no more than 3 terms, numerator degree less than denominator) (refer to the official specification document for exact wording). Although this strand sits in section B, it is examined synoptically across all three papers. Partial fractions underpin integration of rational functions (where $\int \dfrac{1}{x+a}\,dx = \ln|x+a| + C$∫x+a1​dx=ln∣x+a∣+C becomes available only after decomposition), the binomial general expansion for $(1+x)^n$(1+x)n with non-integer $n$n (where $\dfrac{1}{(1-2x)(1+x)}$(1−2x)(1+x)1​ must first be split before each piece is expanded), and first-order linear differential equations with rational coefficients. The AQA formula booklet does not list the partial-fraction patterns — they must be memorised.

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Worked example with full mark scheme

Question (8 marks): Express $f(x) = \dfrac{2x^3 + 5x^2 + 4x + 3}{(x+1)^2(x+2)}$f(x)=(x+1)2(x+2)2x3+5x2+4x+3​ in the form $A + \dfrac{B}{x+1} + \dfrac{C}{(x+1)^2} + \dfrac{D}{x+2}$A+x+1B​+(x+1)2C​+x+2D​, where $A$A, $B$B, $C$C, $D$D are constants to be found.

Solution with mark scheme:

Step 1 — check for improper fraction. The numerator has degree $3$3; the denominator $(x+1)^2(x+2)$(x+1)2(x+2) expands to a cubic, so it also has degree $3$3. The fraction is improper (degree of numerator equals degree of denominator), so polynomial division is required first.

M1 — recognising the improper-fraction structure and committing to division. Common error: launching straight into partial fractions, which produces inconsistent equations.

Step 2 — polynomial division. Expand the denominator: $(x+1)^2(x+2) = (x^2+2x+1)(x+2) = x^3 + 4x^2 + 5x + 2$(x+1)2(x+2)=(x2+2x+1)(x+2)=x3+4x2+5x+2.

Divide $2x^3 + 5x^2 + 4x + 3$2x3+5x2+4x+3 by $x^3 + 4x^2 + 5x + 2$x3+4x2+5x+2:

$2x^3 + 5x^2 + 4x + 3 = 2 \cdot (x^3 + 4x^2 + 5x + 2) + (-3x^2 - 6x - 1)$2x3+5x2+4x+3=2⋅(x3+4x2+5x+2)+(−3x2−6x−1)

So $f(x) = 2 + \dfrac{-3x^2 - 6x - 1}{(x+1)^2(x+2)}$f(x)=2+(x+1)2(x+2)−3x2−6x−1​.

M1 A1 — quotient $A = 2$A=2, correct remainder polynomial.

Step 3 — decompose the proper fraction. Write

$\dfrac{-3x^2 - 6x - 1}{(x+1)^2(x+2)} = \dfrac{B}{x+1} + \dfrac{C}{(x+1)^2} + \dfrac{D}{x+2}$(x+1)2(x+2)−3x2−6x−1​=x+1B​+(x+1)2C​+x+2D​

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

$-3x^2 - 6x - 1 = B(x+1)(x+2) + C(x+2) + D(x+1)^2$−3x2−6x−1=B(x+1)(x+2)+C(x+2)+D(x+1)2

M1 — correct partial-fraction setup with the $C/(x+1)^2$C/(x+1)2 repeated-factor term included.

Step 4 — substitute strategic values (cover-up / substitution).

Let $x = -1$x=−1: $-3 + 6 - 1 = 0 + C(1) + 0$−3+6−1=0+C(1)+0, so $C = 2$C=2.

Let $x = -2$x=−2: $-12 + 12 - 1 = 0 + 0 + D(1)$−12+12−1=0+0+D(1), so $D = -1$D=−1.

M1 — using the cover-up substitution at the roots to extract $C$C and $D$D directly.

Step 5 — find $B$B. Compare $x^2$x2 coefficients: $-3 = B + D$−3=B+D, so $B = -3 - D = -3 - (-1) = -2$B=−3−D=−3−(−1)=−2.

A1 — $B = -2$B=−2. (Equivalent: substitute any third value, e.g. $x = 0$x=0, and solve.)

Step 6 — final answer.

$f(x) = 2 - \dfrac{2}{x+1} + \dfrac{2}{(x+1)^2} - \dfrac{1}{x+2}$f(x)=2−x+12​+(x+1)22​−x+21​

A1 — fully assembled answer with all four constants identified.

Total: 8 marks (M4 A4).

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Specimen question modelled on the AQA 7357 Paper 2 format

Question (6 marks): $g(x) = \dfrac{7x - 5}{(2x - 1)(x + 3)}$g(x)=(2x−1)(x+3)7x−5​.

(a) Express $g(x)$g(x) in partial fractions. (3)

(b) Hence find $\int g(x)\,dx$∫g(x)dx. (3)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — writing $\dfrac{7x-5}{(2x-1)(x+3)} = \dfrac{P}{2x-1} + \dfrac{Q}{x+3}$(2x−1)(x+3)7x−5​=2x−1P​+x+3Q​ and multiplying through.
• M1 (AO1.1b) — cover-up at $x = \tfrac{1}{2}$x=21​: numerator becomes $\tfrac{7}{2} - 5 = -\tfrac{3}{2}$27​−5=−23​, denominator factor $(x+3)$(x+3) evaluates to $\tfrac{7}{2}$27​, giving $P = -\tfrac{3}{2} \div \tfrac{7}{2} = -\tfrac{3}{7}$P=−23​÷27​=−73​. Cover-up at $x = -3$x=−3: $-21 - 5 = -26$−21−5=−26, divided by $(2(-3)-1) = -7$(2(−3)−1)=−7, giving $Q = \tfrac{26}{7}$Q=726​.
• A1 (AO1.1b) — $g(x) = -\dfrac{3/7}{2x-1} + \dfrac{26/7}{x+3}$g(x)=−2x−13/7​+x+326/7​, equivalently $\dfrac{1}{7}\left(-\dfrac{3}{2x-1} + \dfrac{26}{x+3}\right)$71​(−2x−13​+x+326​).

(b)

• M1 (AO1.1b) — applying $\int \dfrac{1}{ax+b}\,dx = \tfrac{1}{a}\ln|ax+b| + C$∫ax+b1​dx=a1​ln∣ax+b∣+C to each piece.
• A1 (AO1.1b) — $-\tfrac{3}{14}\ln|2x-1| + \tfrac{26}{7}\ln|x+3| + C$−143​ln∣2x−1∣+726​ln∣x+3∣+C.
• A1 (AO2.5) — modulus signs present and constant of integration included.

Total: 6 marks split AO1 = 5, AO2 = 1. AQA examiners typically reserve the AO2 mark on partial-fractions questions for presentation craft: modulus signs on logarithms, the explicit $+C$+C, and the linear-coefficient correction $\tfrac{1}{a}$a1​ in the integral.

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Synoptic links

Connects to:

1. Integration of rational functions: the entire purpose of partial-fraction decomposition at A-Level is to convert $\int \dfrac{P(x)}{Q(x)}\,dx$∫Q(x)P(x)​dx into a sum of $\int \dfrac{1}{x+a}\,dx$∫x+a1​dx and $\int \dfrac{1}{(x+a)^2}\,dx$∫(x+a)21​dx pieces, each of which has a closed form ($\ln|x+a|$ln∣x+a∣ and $-\dfrac{1}{x+a}$−x+a1​ respectively). Without decomposition, integrals like $\int \dfrac{1}{x^2 - 1}\,dx$∫x2−11​dx are inaccessible.
2. Binomial general expansion ($(1+x)^n$(1+x)n for rational $n$n): to expand $\dfrac{1}{(1-2x)(1+x)}$(1−2x)(1+x)1​ as a power series in $x$x, decompose into $\dfrac{A}{1-2x} + \dfrac{B}{1+x}$1−2xA​+1+xB​ first; each piece is then expanded as a binomial series valid for $|2x| < 1$∣2x∣<1 and $|x| < 1$∣x∣<1 respectively, and the series are summed term-by-term. The radius of convergence is the minimum of the two.
3. First-order linear differential equations with separation of variables: equations of the form $\dfrac{dy}{dx} = \dfrac{f(x)}{g(y)}$dxdy​=g(y)f(x)​ where $g(y)$g(y) factorises into linear factors require partial-fraction decomposition on the $y$y-side before integrating. Logistic-growth equations $\dfrac{dP}{dt} = kP(M - P)$dtdP​=kP(M−P) are the canonical example.
4. Improper integrals and limits at infinity: integrating $\dfrac{1}{(x+1)(x+2)}$(x+1)(x+2)1​ from $0$0 to $\infty$∞ exploits the partial-fraction form to combine logarithms before taking the limit, since $\ln\left(\dfrac{x+1}{x+2}\right) \to \ln(1) = 0$ln(x+2x+1​)→ln(1)=0 as $x \to \infty$x→∞.
5. Numerical methods (Year 2 section P): when partial fractions decompose into pieces whose integrals are inaccessible (e.g. $\dfrac{1}{x^2 + 1}$x2+11​ requires $\arctan$arctan, beyond AQA), numerical integration via the trapezium rule or Simpson's rule provides the fallback.

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Mark-scheme literacy

Partial-fractions questions on AQA 7357 split AO marks heavily toward AO1, with AO2 reserved for the connective tissue between decomposition and the synoptic application:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	65–75%	Recognising improper fractions and dividing first; setting up the correct partial-fraction template with repeated factors; applying cover-up; solving for unknown constants
AO2 (reasoning / interpretation)	20–30%	Justifying the choice of partial-fraction template (why $A/(x+1) + B/(x+1)^2$A/(x+1)+B/(x+1)2 rather than $A/(x+1)^2$A/(x+1)2 alone); presenting answers consistent with the form requested; modulus signs on logarithms
AO3 (problem-solving)	0–10%	Multi-step contexts: partial fractions \to\ integration \to\ definite integral with limits; appears mostly on Paper 3

Examiner-rewarded phrasing: "since the numerator has degree at least as large as the denominator, polynomial division is required first"; "for the repeated factor $(x+1)^2$(x+1)2, the decomposition includes both $A/(x+1)$A/(x+1) and $B/(x+1)^2$B/(x+1)2"; "by the cover-up rule, evaluating at $x = -1$x=−1 isolates the $(x+1)^2$(x+1)2 coefficient". Phrases that lose marks: "let $x = -1$x=−1" without justifying why this kills three of the four unknowns; setting up only $A/(x+1)^2$A/(x+1)2 for a repeated factor (missing the $B/(x+1)$B/(x+1) term); writing $\ln(x+1)$ln(x+1) without the modulus where the domain is unrestricted.

A specific AQA pattern: questions that say "Hence find $\int g(x)\,dx$∫g(x)dx" mean use the partial fractions you just computed. Restarting from scratch with a substitution method earns no method marks because the "Hence" command is binding.

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Grade-band model answers

3-mark question

Question: Express $\dfrac{5x + 1}{(x - 1)(x + 2)}$(x−1)(x+2)5x+1​ in partial fractions.

Grade C response (~180 words):

Let $\dfrac{5x+1}{(x-1)(x+2)} = \dfrac{A}{x-1} + \dfrac{B}{x+2}$(x−1)(x+2)5x+1​=x−1A​+x+2B​. Multiply through:

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

Let $x = 1$x=1: $6 = 3A$6=3A, so $A = 2$A=2.

Let $x = -2$x=−2: $-9 = -3B$−9=−3B, so $B = 3$B=3.

Answer: $\dfrac{2}{x-1} + \dfrac{3}{x+2}$x−12​+x+23​.

Examiner commentary: Full marks (3/3). The candidate identifies the partial-fraction template, multiplies through correctly, and uses cover-up substitutions at the roots to extract both constants. The working is brief but every step is verifiable. Many Grade C candidates lose a mark by setting up a system of two equations in two unknowns instead of using cover-up — this works algebraically but consumes time and increases arithmetic-error risk. Choosing strategic substitutions $x = 1$x=1 and $x = -2$x=−2 (the roots) is the efficient method examiners reward.

Grade A response (~240 words):*

Set up the partial-fraction decomposition:

$\dfrac{5x + 1}{(x - 1)(x + 2)} = \dfrac{A}{x - 1} + \dfrac{B}{x + 2}$(x−1)(x+2)5x+1​=x−1A​+x+2B​

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

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

By the cover-up method, substitute $x = 1$x=1 (the root of the first factor): $5(1) + 1 = A(3) + 0$5(1)+1=A(3)+0, giving $A = 2$A=2. Substitute $x = -2$x=−2 (the root of the second factor): $5(-2) + 1 = 0 + B(-3)$5(−2)+1=0+B(−3), giving $-9 = -3B$−9=−3B, so $B = 3$B=3.

Therefore:

$\dfrac{5x + 1}{(x - 1)(x + 2)} = \dfrac{2}{x - 1} + \dfrac{3}{x + 2}$(x−1)(x+2)5x+1​=x−12​+x+23​

Examiner commentary: Full marks (3/3). Beyond procedural correctness, the candidate names the technique ("cover-up method") and explains why the chosen substitutions are strategic — they kill all but one term. Verification by re-combining the right-hand side over a common denominator (not shown but available) would push the answer beyond examination craft into mathematical maturity. The display-math notation and the explicit linkage of each substitution to the corresponding root is the kind of sign-posting that signals examiner-aware sophistication.

6-mark question

Question: Express $\dfrac{4x^2 + x + 3}{(x+1)(x^2 + 1)}$(x+1)(x2+1)4x2+x+3​ in partial fractions.

Grade B response (~270 words):

Set $\dfrac{4x^2 + x + 3}{(x+1)(x^2+1)} = \dfrac{A}{x+1} + \dfrac{Bx + C}{x^2+1}$(x+1)(x2+1)4x2+x+3​=x+1A​+x2+1Bx+C​.

Multiply: $4x^2 + x + 3 = A(x^2 + 1) + (Bx + C)(x + 1)$4x2+x+3=A(x2+1)+(Bx+C)(x+1).

Let $x = -1$x=−1: $4 - 1 + 3 = 2A$4−1+3=2A, so $A = 3$A=3.

Compare $x^2$x2: $4 = A + B = 3 + B$4=A+B=3+B, so $B = 1$B=1.

Compare constants: $3 = A + C = 3 + C$3=A+C=3+C, so $C = 0$C=0.

Answer: $\dfrac{3}{x+1} + \dfrac{x}{x^2+1}$x+13​+x2+1x​.

Examiner commentary: Method correct throughout, secured 5/6. The candidate sets up the irreducible-quadratic template ($Bx + C$Bx+C over $x^2 + 1$x2+1) correctly, applies cover-up at $x = -1$x=−1, and uses coefficient comparison for the remaining unknowns. The lost mark is for not verifying — re-combining $\dfrac{3}{x+1} + \dfrac{x}{x^2+1}$x+13​+x2+1x​ over the common denominator should yield $\dfrac{4x^2 + x + 3}{(x+1)(x^2+1)}$(x+1)(x2+1)4x2+x+3​, and a brief check would have caught any arithmetic slip. Note that AQA 7357 only examines decompositions where the irreducible quadratic factor is explicitly given; candidates need not factorise quadratics into complex linear pieces.

Grade A response (~300 words):*

The denominator factors as $(x+1)(x^2+1)$(x+1)(x2+1), where $x^2 + 1$x2+1 is irreducible over the reals. The partial-fraction template therefore takes the form:

$\dfrac{4x^2 + x + 3}{(x+1)(x^2+1)} = \dfrac{A}{x+1} + \dfrac{Bx + C}{x^2+1}$(x+1)(x2+1)4x2+x+3​=x+1A​+x2+1Bx+C​

Multiplying both sides by $(x+1)(x^2+1)$(x+1)(x2+1):

$4x^2 + x + 3 = A(x^2 + 1) + (Bx + C)(x + 1)$4x2+x+3=A(x2+1)+(Bx+C)(x+1).

Cover-up at $x = -1$x=−1: $4(1) + (-1) + 3 = A(2)$4(1)+(−1)+3=A(2), so $2A = 6$2A=6 and $A = 3$A=3.

Compare $x^2$x2 coefficients: $4 = A + B = 3 + B$4=A+B=3+B, so $B = 1$B=1.

Compare constant terms: $3 = A + C = 3 + C$3=A+C=3+C, so $C = 0$C=0.

Therefore:

$\dfrac{4x^2 + x + 3}{(x+1)(x^2+1)} = \dfrac{3}{x + 1} + \dfrac{x}{x^2 + 1}$(x+1)(x2+1)4x2+x+3​=x+13​+x2+1x​

Verification: $\dfrac{3(x^2+1) + x(x+1)}{(x+1)(x^2+1)} = \dfrac{3x^2 + 3 + x^2 + x}{(x+1)(x^2+1)} = \dfrac{4x^2 + x + 3}{(x+1)(x^2+1)}$(x+1)(x2+1)3(x2+1)+x(x+1)​=(x+1)(x2+1)3x2+3+x2+x​=(x+1)(x2+1)4x2+x+3​.

Examiner commentary: Full marks (6/6). The candidate notes the irreducibility of $x^2 + 1$x2+1 explicitly (justifying the linear-numerator template), applies cover-up, and verifies. The verification step is the AO2 mark: it converts a correct answer into an examiner-rewarded answer by demonstrating the candidate has connected the decomposition back to the original.

9-mark question

Question: $h(x) = \dfrac{x^2 + 5x + 6}{(x-1)(x+2)^2}$h(x)=(x−1)(x+2)2x2+5x+6​.

(a) Express $h(x)$h(x) in partial fractions. (5)

(b) Hence find $\int_2^3 h(x)\,dx$∫23​h(x)dx, giving your answer in the form $\ln k + m$lnk+m with $k$k, $m$m rational. (4)

Grade A response (~360 words):*

(a) Set $\dfrac{x^2 + 5x + 6}{(x-1)(x+2)^2} = \dfrac{A}{x-1} + \dfrac{B}{x+2} + \dfrac{C}{(x+2)^2}$(x−1)(x+2)2x2+5x+6​=x−1A​+x+2B​+(x+2)2C​.

Multiply: $x^2 + 5x + 6 = A(x+2)^2 + B(x-1)(x+2) + C(x-1)$x2+5x+6=A(x+2)2+B(x−1)(x+2)+C(x−1).

Cover-up at $x = 1$x=1: $1 + 5 + 6 = 9A$1+5+6=9A, so $A = \tfrac{12}{9} = \tfrac{4}{3}$A=912​=34​.

Cover-up at $x = -2$x=−2: $4 - 10 + 6 = -3C$4−10+6=−3C, so $0 = -3C$0=−3C, giving $C = 0$C=0.

Compare $x^2$x2: $1 = A + B = \tfrac{4}{3} + B$1=A+B=34​+B, so $B = -\tfrac{1}{3}$B=−31​.

So $h(x) = \dfrac{4/3}{x-1} - \dfrac{1/3}{x+2}$h(x)=x−14/3​−x+21/3​ (the $C/(x+2)^2$C/(x+2)2 term vanishes here).

(b) $\int_2^3 h(x)\,dx = \left[\tfrac{4}{3}\ln|x-1| - \tfrac{1}{3}\ln|x+2|\right]_2^3$∫23​h(x)dx=[34​ln∣x−1∣−31​ln∣x+2∣]23​.

At $x = 3$x=3: $\tfrac{4}{3}\ln 2 - \tfrac{1}{3}\ln 5$34​ln2−31​ln5. At $x = 2$x=2: $\tfrac{4}{3}\ln 1 - \tfrac{1}{3}\ln 4 = 0 - \tfrac{2}{3}\ln 2$34​ln1−31​ln4=0−32​ln2.

Subtract: $\tfrac{4}{3}\ln 2 - \tfrac{1}{3}\ln 5 - (-\tfrac{2}{3}\ln 2) = 2\ln 2 - \tfrac{1}{3}\ln 5$34​ln2−31​ln5−(−32​ln2)=2ln2−31​ln5.

Combine: $2\ln 2 = \ln 4$2ln2=ln4, so the answer is $\ln 4 - \tfrac{1}{3}\ln 5$ln4−31​ln5, equivalently $\ln\left(\dfrac{4}{5^{1/3}}\right)$ln(51/34​).

Examiner commentary: Full marks (9/9). The candidate spots that $C = 0$C=0 (the repeated-factor coefficient vanishes for this particular numerator) and reports it explicitly rather than discarding the term silently. In (b), modulus signs are present, limits are evaluated cleanly, and the final form is presented in two equivalent presentations — the additive form $\ln 4 - \tfrac{1}{3}\ln 5$ln4−31​ln5 and the consolidated logarithm $\ln(4/5^{1/3})$ln(4/51/3) — leaving the examiner to award the AO2 mark for either.

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A-Level-depth misconceptions

The errors that distinguish A from A* on partial-fractions questions:

1. Forgetting the repeated-factor pattern. A factor $(x+1)^2$(x+1)2 in the denominator demands two terms in the decomposition: $\dfrac{A}{x+1} + \dfrac{B}{(x+1)^2}$x+1A​+(x+1)2B​. Writing only $\dfrac{B}{(x+1)^2}$(x+1)2B​ produces an underdetermined system that appears to work for cover-up but fails coefficient comparison.

2. Missing the degree-of-numerator check. Partial-fraction decomposition is only valid for proper fractions (numerator degree strictly less than denominator). If degrees are equal or the numerator is larger, polynomial division is required first. Skipping this step produces inconsistent simultaneous equations.

3. Cover-up at a non-root. The cover-up method works because substituting $x = a$x=a kills every term containing $(x - a)$(x−a) as a factor. Substituting any other value produces an equation in all unknowns, not just one.

4. Irreducible-quadratic template. A factor like $x^2 + 1$x2+1 in the denominator demands a linear numerator $Bx + C$Bx+C, not a constant. Writing $\dfrac{B}{x^2 + 1}$x2+1B​ alone underdetermines the system.

5. Modulus on logarithms. When integrating $\dfrac{1}{x+a}$x+a1​ over a domain that includes both signs of $x + a$x+a, the antiderivative is $\ln|x+a|$ln∣x+a∣, not $\ln(x+a)$ln(x+a). AQA examiners deduct the final A1 for missing modulus signs.

6. Linear-coefficient correction in integration. $\int \dfrac{1}{2x - 1}\,dx = \tfrac{1}{2}\ln|2x - 1| + C$∫2x−11​dx=21​ln∣2x−1∣+C, not $\ln|2x - 1| + C$ln∣2x−1∣+C. The $\tfrac{1}{a}$a1​ from the chain rule is the most-missed factor across all integration questions.

7. Sign errors in cover-up at negative roots. Substituting $x = -2$x=−2 into $(x+2)$(x+2) gives $0$0, but substituting into $(x-1)$(x−1) gives $-3$−3, not $3$3. Sign-management at negative roots costs a mark roughly half the time.

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Common errors and mark-loss patterns on partial fractions

Several patterns repeatedly cost candidates marks on Paper 2 partial-fractions questions. They are all about structural awareness, not arithmetic.

• Skipping the improper-fraction check. When the numerator has degree at least as large as the denominator, partial fractions cannot proceed without polynomial division first. Candidates who launch straight into setting up $A/(x+1) + B/(x+2)$A/(x+1)+B/(x+2) produce equations that have no consistent solution — and the time spent debugging them is time lost on the rest of the paper.
• Conflating repeated-factor templates with single-factor templates. A repeated linear factor demands both the $A/(x+a)$A/(x+a) and $B/(x+a)^2$B/(x+a)2 pieces. Many candidates write only the higher-power term, then panic when coefficient comparison produces contradictions. The cure: write the template immediately on reading the denominator, before any algebra.
• Mis-identifying which substitution kills which term. Cover-up works at the roots of the denominator. For $(x-1)(x+2)^2$(x−1)(x+2)2, substituting $x = 1$x=1 isolates the $A/(x-1)$A/(x−1) coefficient and substituting $x = -2$x=−2 isolates the $C/(x+2)^2$C/(x+2)2 coefficient — but $B$B requires either coefficient comparison or a third substitution.
• Dropping the $+C$+C on indefinite integrals derived from partial fractions. Even when the partial-fraction decomposition is perfect, a missing constant of integration costs the final accuracy mark on every "find $\int f(x)\,dx$∫f(x)dx" question. Examiners check this ruthlessly.

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Going further — university and academic signposting

Partial-fraction decomposition points directly toward several undergraduate trajectories:

• Complex analysis (Year 2/3): the residue theorem generalises partial-fraction decomposition to meromorphic functions over the complex plane. The "cover-up" coefficients you compute at A-Level are the residues at simple poles, and contour integration of $\oint \dfrac{f(z)}{z - a}\,dz$∮z−af(z)​dz reduces to $2\pi i$2πi times the residue at $z = a$z=a.
• Linear ODEs and Laplace transforms (Year 1/2 engineering and applied maths): solving constant-coefficient linear ODEs via Laplace transforms reduces to inverting a rational function in the Laplace variable $s$s, which is exactly partial-fraction decomposition followed by table lookup. Every textbook circuit-analysis problem rests on this reduction.
• Algebraic geometry (Year 3+): the structure theorem for finitely generated modules over a principal ideal domain — which classifies abelian groups, finite-dimensional vector spaces, and Jordan canonical form — generalises the "every rational function decomposes into a sum of simple fractions" theorem to a vastly more abstract setting.
• Generating functions in combinatorics (Year 2): counting problems whose generating functions are rational reduce, via partial fractions, to closed-form expressions for the $n$nth term — the Fibonacci closed form $F_n = \tfrac{1}{\sqrt{5}}(\phi^n - \psi^n)$Fn​=5​1​(ϕn−ψn) is derived this way from $\dfrac{x}{1 - x - x^2}$1−x−x2x​.

Oxbridge interview prompt: "Decompose $\dfrac{1}{x^2 - 1}$x2−11​ into partial fractions, then evaluate $\sum_{n=2}^{\infty} \dfrac{1}{n^2 - 1}$∑n=2∞​n2−11​ exactly. Why does the sum telescope?"

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Additional A* practice: integrating partial fractions

A common A* trap on AQA 7357 Paper 2 is to chain partial-fraction decomposition with definite integration and to require an exact answer in logarithm form.

Worked example: Find $\displaystyle\int \dfrac{3x + 5}{(x+1)(x-2)}\,dx$∫(x+1)(x−2)3x+5​dx.

Step 1 — decompose. Write $\dfrac{3x+5}{(x+1)(x-2)} = \dfrac{A}{x+1} + \dfrac{B}{x-2}$(x+1)(x−2)3x+5​=x+1A​+x−2B​.

Multiply: $3x + 5 = A(x - 2) + B(x + 1)$3x+5=A(x−2)+B(x+1).

Cover-up at $x = -1$x=−1: $-3 + 5 = A(-3) + 0$−3+5=A(−3)+0, so $2 = -3A$2=−3A, giving $A = -\tfrac{2}{3}$A=−32​.

Cover-up at $x = 2$x=2: $6 + 5 = 0 + 3B$6+5=0+3B, so $B = \tfrac{11}{3}$B=311​.

Step 2 — integrate term by term.

$\int \dfrac{3x + 5}{(x+1)(x-2)}\,dx = -\tfrac{2}{3}\int \dfrac{1}{x+1}\,dx + \tfrac{11}{3}\int \dfrac{1}{x-2}\,dx$∫(x+1)(x−2)3x+5​dx=−32​∫x+11​dx+311​∫x−21​dx

Each piece integrates to a logarithm:

$= -\tfrac{2}{3}\ln|x+1| + \tfrac{11}{3}\ln|x-2| + C$=−32​ln∣x+1∣+311​ln∣x−2∣+C

Step 3 — combine (optional consolidation). Using log laws:

$= \tfrac{1}{3}\ln\left|\dfrac{(x-2)^{11}}{(x+1)^2}\right| + C$=31​ln​(x+1)2(x−2)11​​+C

Why A candidates spot this immediately:* the structure "rational function with linear factors in the denominator" is the signature of a partial-fractions integration. Every time you see $\int \dfrac{P(x)}{Q(x)}\,dx$∫Q(x)P(x)​dx with $Q$Q factorising into linear pieces and $\deg P < \deg Q$degP<degQ, decompose first. The same trick handles $\int \dfrac{2x-1}{x^2 - 4}\,dx$∫x2−42x−1​dx (factor $x^2 - 4 = (x-2)(x+2)$x2−4=(x−2)(x+2)) and $\int \dfrac{1}{x(x+1)}\,dx$∫x(x+1)1​dx (decompose to $\dfrac{1}{x} - \dfrac{1}{x+1}$x1​−x+11​, integrate to $\ln|x| - \ln|x+1| = \ln\left|\dfrac{x}{x+1}\right|$ln∣x∣−ln∣x+1∣=ln​x+1x​​).

A subtlety: when the denominator contains a repeated factor, the decomposition produces a $\dfrac{B}{(x+a)^2}$(x+a)2B​ piece whose integral is $-\dfrac{B}{x+a}$−x+aB​, not a logarithm. Mixing the two integration patterns is the most-frequent slip on chained questions.

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AQA A-Level alignment footer

This content is aligned with the AQA A-Level Mathematics (7357) specification, Paper 2 — Pure Mathematics, Section B: Algebra and Functions. For the most accurate and up-to-date information, please refer to the official AQA specification document.

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Visual summary

graph TD
    A["Rational function<br/>P(x)/Q(x)"] --> B{"Is deg numerator < deg denominator?"}
    B -->|"No (improper)"| C["Polynomial division<br/>first: P/Q = A + R/Q"]
    B -->|"Yes (proper)"| D{"Factor structure<br/>of denominator?"}
    C --> D
    D -->|"Distinct linear"| E["Template:<br/>A/(x+a) + B/(x+b)<br/>Cover-up at roots"]
    D -->|"Repeated linear"| F["Template:<br/>A/(x+a) + B/(x+a)² + C/(x+b)<br/>Cover-up plus compare"]
    E --> G["Integrate:<br/>each piece to logarithm"]
    F --> G

    style C fill:#27ae60,color:#fff
    style G fill:#3498db,color:#fff