Partial Fractions — Decomposition and Integration — Pure Mathematics 4

Before you read on

Two or three questions on exactly what this lesson teaches. Being wrong here is fine — it's the fastest way to find out what to pay attention to next.

What partial-fraction decomposition is, and why this paper pairs it with integration

Spec 2.1's own guidance is precise about the shape of denominator this course covers: "Partial fractions to include denominators such as $(ax+b)(cx+d)(ex+f)$ and $(ax+b)(cx+d)^2$ " — three distinct linear factors, or one linear factor alongside a repeated one — "The degree of the numerator may equal or exceed the degree of the denominator" (a required case, not an edge case), and explicitly: quadratic factors like $(x^2+a)$ , $a>0$ , "are not required." The same guidance line states the point of learning any of this at all: "Applications to integration, differentiation and series expansions."

Spec 6.3 is where that first application actually lands: "Simple cases of integration using partial fractions." Its own guidance names the exact shapes: "Integration of rational expressions such as those arising from partial fractions, e.g. $\frac{2}{3x+5}$ , $\frac{3}{(x-1)^2}$ ." A with a genuinely mixed denominator — several different linear factors, or a repeated one — usually has NO direct you could just write down. Split into partial fractions, though, every single term becomes one of exactly two familiar shapes: $\frac{1}{ax+b}$ , which integrates to a , or $\frac{1}{(ax+b)^n}$ for $n\geq 2$ , which integrates by the reverse power rule instead. Decomposition doesn't just simplify the algebra — it is the ONLY reason the integral becomes tractable at all. This lesson works through the decomposition half first (spec 2.1), then the integration it feeds (spec 6.3), in exactly the order the spec's own guidance line describes.

Mechanism

Why a partial-fraction identity can be checked at chosen values of x — and why matching coefficients works too

A partial-fraction decomposition is written with the symbol ≡, not =, and that distinction is the entire justification for both standard solving methods. An equation like $2x=6$ is only true for one value of $x$ ; an like $8x^2+8x-3\equiv A(x+1)(2x+1)+B(x-2)(2x+1)+C(x-2)(x+1)$ is claimed to be true for every single value of $x$ at once — which means it is safe to substitute ANY value into it, including a value that makes one of the original fraction's own denominators zero (the identity itself, once cleared of denominators, is an ordinary polynomial equation defined everywhere, even where the original fraction wasn't). Substituting the root of one bracket makes every OTHER bracket's term vanish on the right-hand side, leaving an equation in only one unknown — this is the cover-up rule, and it works precisely because the identity is guaranteed to hold at that value too, not despite it. A second, completely independent consequence of the same ≡ symbol: two polynomials that are equal for every value of $x$ must have identical coefficients of every power of $x$ — $x^2$ , $x^1$ , $x^0$ separately — since a polynomial is entirely determined by its coefficients. This is why comparing coefficients is an equally valid, equally rigorous route to the same unknowns, not an approximation or a shortcut: both methods are reading off different, but equally true, consequences of one ≡ statement.

Worked, in full

Decompose $\dfrac{8x^2+8x-3}{(x-2)(x+1)(2x+1)}$ — three distinct linear factors, spec 2.1's own named shape $(ax+b)(cx+d)(ex+f)$

01
Set up the identity with one unknown per factor: $\dfrac{8x^2+8x-3}{(x-2)(x+1)(2x+1)}\equiv\dfrac{A}{x-2}+\dfrac{B}{x+1}+\dfrac{C}{2x+1}$ . Multiply both sides by the full denominator to clear every fraction: $8x^2+8x-3\equiv A(x+1)(2x+1)+B(x-2)(2x+1)+C(x-2)(x+1)$ .
Earns: M1 — sets up the correct partial-fraction form (one unknown per distinct linear factor) and clears denominators correctly.
02
Substitute $x=2$ (the root of $x-2$ ) — this makes the $B$ and $C$ terms vanish entirely, since both still contain the factor $(x-2)$ : $8(4)+16-3=A(3)(5)+0+0\Rightarrow 45=15A\Rightarrow A=3$ .
Earns: A1 — correctly isolates and solves for A using the cover-up substitution.
03
Substitute $x=-1$ (the root of $x+1$ ) — the $A$ and $C$ terms vanish: $8(1)-8-3=0+B(-3)(-1)+0\Rightarrow -3=3B\Rightarrow B=-1$ .
Earns: A1 — correctly isolates and solves for B.
04
Substitute $x=-\frac{1}{2}$ (the root of $2x+1$ ) — the $A$ and $B$ terms vanish: $8\left(\frac{1}{4}\right)-4-3=0+0+C\left(-\frac{5}{2}\right)\left(\frac{1}{2}\right)\Rightarrow -5=-\frac{5}{4}C\Rightarrow C=4$ .
Earns: A1 — correctly isolates and solves for C, including handling the fractional root cleanly.
05
Check at a fourth, unused value — $x=0$ : left side $\dfrac{-3}{(-2)(1)(1)}=\dfrac{3}{2}$ ; right side $\dfrac{3}{-2}-\dfrac{-1}{1}+\dfrac{4}{1}=-1.5+1+4=1.5$ . ✓ Final answer: $\dfrac{8x^2+8x-3}{(x-2)(x+1)(2x+1)}\equiv\dfrac{3}{x-2}-\dfrac{1}{x+1}+\dfrac{4}{2x+1}$ .
Earns: Nothing — not a markable step, but checking at a value none of the three cover-up substitutions used is the one test that would catch an arithmetic slip in any of A, B or C before it costs marks two blocks from now, when this exact decomposition gets integrated.

Mechanism

Why a repeated factor needs two terms, not one

This is genuine mathematical reasoning about why the form works, not a sourced examiner finding — it is included because it is checkable in a few lines and explains a rule this course would otherwise just assert. Take a denominator of degree 3 built from one linear factor and one repeated one, like $(x+2)(3x-1)^2$ . A genuinely PROPER fraction over this denominator can have any numerator of degree up to 2 — that's three independent numbers to match: the coefficients of $x^2$ , $x^1$ and $x^0$ . Whatever sits on the right of the ≡ sign has to carry enough independent unknowns to be ABLE to match any such numerator, not just one convenient one. $\frac{A}{x+2}$ contributes exactly one unknown. If the repeated factor contributed only $\frac{B}{(3x-1)^2}$ — a single term — the whole right-hand side would carry just two unknowns total ( $A$ and $B$ ), enough to match a numerator with only two independent coefficients, never a general one with three. Splitting the repeated factor into TWO terms, $\frac{B}{3x-1}+\frac{C}{(3x-1)^2}$ , adds the second unknown the count actually needs — three unknowns for three coefficients, exactly matching what a general proper numerator of this degree requires. The rule isn't a memorised pattern; it's the smallest number of terms that could ever work.

Same question, every valid method

Express $\dfrac{11x^2-26x+1}{(3x+1)(x-2)^2}$ in partial fractions. VERIDIAN-original numeric scenario, built around spec 2.1's other named denominator shape $(ax+b)(cx+d)^2$ — deliberately using a different repeated bracket, $(x-2)^2$ , from the real $(2x+1)^2$ -shaped question the trap taxonomy below cites, so the two are never confused with each other.

2 valid methods · every one reaches $\dfrac{2}{3x+1}+\dfrac{3}{x-2}-\dfrac{1}{(x-2)^2}$ · 5 marks available

01
Set up and clear denominators: $11x^2-26x+1\equiv A(x-2)^2+B(3x+1)(x-2)+C(3x+1)$ .
Method mark for the correct form (one term each for the single factor and the two powers of the repeated factor) and correctly cleared denominators.
M1
02
Substitute $x=2$ (the repeated root — this vanishes both the $A$ and $B$ terms at once): $11(4)-52+1=C(7)\Rightarrow -7=7C\Rightarrow C=-1$ .
Accuracy mark for C, using the repeated root to isolate the HIGHEST power's own coefficient directly — the one substitution that removes two unknowns at once, not just one.
A1
03
Substitute $x=-\frac{1}{3}$ (root of $3x+1$ ): $11\left(\frac{1}{9}\right)+\frac{26}{3}+1=A\left(-\frac{7}{3}\right)^2\Rightarrow \frac{98}{9}=\frac{49}{9}A\Rightarrow A=2$ .
Accuracy mark for A.
A1
04
Substitute a third, convenient value — $x=0$ — now that $A$ and $C$ are both known: $1=A(4)+B(1)(-2)+C(1)=8-2B-1\Rightarrow 2B=6\Rightarrow B=3$ .
Accuracy mark for B — the one constant the repeated-root and single-root substitutions can't reach directly, found here from a third value rather than by expanding and comparing coefficients.
A1
05
$\dfrac{11x^2-26x+1}{(3x+1)(x-2)^2}\equiv\dfrac{2}{3x+1}+\dfrac{3}{x-2}-\dfrac{1}{(x-2)^2}$ .
Accuracy mark (cao) for correctly stating the complete final partial-fraction decomposition using the values of A, B and C found above: 2/(3x+1) + 3/(x-2) - 1/(x-2)².
A1

A real examiner report on a repeated-linear-factor decomposition question records that simultaneous equations built from expanded coefficients were used by a real, significant number of candidates but were "more prone to error than attempts via substitution" (Jan 2024, Q2(a)) — this method is the one the real record shows working best in practice. It also needs no expansion of brackets at all for two of the three unknowns.

When the numerator's degree isn't smaller than the denominator's

Spec 2.1 states this directly, as required content rather than an edge case: "The degree of the numerator may equal or exceed the degree of the denominator." The basic form $\frac{A}{x-a}+\frac{B}{x-b}$ cannot represent every fraction with this denominator — each term individually tends to 0 as $x\to\infty$ , so their sum can never reproduce a fraction that settles to a nonzero constant, or grows without bound, as $x$ gets large. A fraction whose numerator's degree matches or exceeds its denominator's does exactly that, so the basic form has to be extended before it can work at all.

Two genuinely different, both legitimate routes fix this, and a real examiner report on exactly this kind of question treats both as equally valid (Oct 2021, Q3(a)). The first: , dividing the numerator by the denominator first to extract a polynomial quotient, leaving behind a genuinely proper remainder over the original denominator — decompose that remainder normally. The second: the identity method, building the polynomial part directly into the assumed form from the start (for example, $\equiv D+\frac{A}{x-a}+\frac{B}{x-b}$ when the degree gap is exactly one) and solving every unknown, including $D$ , from one cleared identity at once — usually by matching the highest power's coefficient first, since only the polynomial part contributes to it.

Marked, line by line

Express $\dfrac{2x^2+4x+3}{(x-1)(x+2)}$ in partial fractions, using long division to find the polynomial part first. (5) VERIDIAN-original scenario: numerator and denominator have equal degree (2), the case spec 2.1 explicitly requires. Both the numbers and the mark tariff below are original, modelled on verified mark-scheme conventions (§4 of the facts bank) rather than transcribed from a real scheme, which for an original question does not exist.

5 marks available

(a) divide — 2 marks

01
Denominator expanded: $(x-1)(x+2)=x^2+x-2$ . Divide: the leading terms give quotient term $2$ ( $2x^2\div x^2$ ). Multiply back: $2(x^2+x-2)=2x^2+2x-4$ . Subtract: $(2x^2+4x+3)-(2x^2+2x-4)=2x+7$ .
Method mark for attempting the division and correctly subtracting to find the remainder — since numerator and denominator have equal degree, the quotient is a constant, leaving a proper-fraction remainder one degree lower.
M1
02
So $\dfrac{2x^2+4x+3}{(x-1)(x+2)}=2+\dfrac{2x+7}{(x-1)(x+2)}$ .
Accuracy mark for the correctly stated quotient-plus-remainder identity.
A1

(b) decompose the remainder — 3 marks

101
Set up: $2x+7\equiv A(x+2)+B(x-1)$ .
Method mark for the correct partial-fraction form applied to the proper remainder from part (a), and correctly cleared denominators.
M1
102
Substitute $x=1$ : $9=3A\Rightarrow A=3$ . Substitute $x=-2$ : $3=-3B\Rightarrow B=-1$ .
Accuracy mark for both A and B, correctly found by cover-up substitution.
A1
103
$\dfrac{2x^2+4x+3}{(x-1)(x+2)}\equiv 2+\dfrac{3}{x-1}-\dfrac{1}{x+2}$ .
Accuracy mark for the complete final decomposition, including the constant term from part (a).
A1

Mechanism

Where the 1/a in ∫1/(ax+b)dx = (1/a)ln|ax+b| + c actually comes from

Let $y=\ln|ax+b|$ . By the (outer function $\ln(u)$ , inner function $u=ax+b$ , $\frac{du}{dx}=a$ ): $\frac{dy}{dx}=\frac{1}{ax+b}\cdot a=\frac{a}{ax+b}$ . So $\int \frac{a}{ax+b}\,dx=\ln|ax+b|+c$ , and dividing both sides by $a$ : $\int \frac{1}{ax+b}\,dx=\frac{1}{a}\ln|ax+b|+c$ . The $\frac{1}{a}$ is not a decoration or a convention to memorise — it exists specifically to cancel the extra factor of $a$ that the chain rule leaves behind when $\ln|ax+b|$ is differentiated. Omitting it is mathematically equivalent to claiming $\ln|ax+b|$ differentiates to exactly $\frac{1}{ax+b}$ — which is only ever true when $a=1$ . For a REPEATED linear factor, $n\geq 2$ , the antiderivative is a completely different family, with no logarithm anywhere in it: reversing the chain rule on $(ax+b)^{1-n}$ gives $\int \frac{1}{(ax+b)^n}\,dx=\frac{-1}{(n-1)a(ax+b)^{n-1}}+c$ . Both results come from running the chain rule backwards — one on $\ln(ax+b)$ , the other on a negative power of $(ax+b)$ — but they are genuinely different antiderivative shapes, not two versions of the same rule, which is exactly why mixing them up (using a logarithm where a power is needed, or vice versa) is a real, checkable error rather than a rounding-level slip.

The rest of spec 6.3's own named guidance

Spec 6.3 names four specific integrands in its own guidance text: $\frac{2}{3x+5}$ , $\frac{3}{(x-1)^2}$ , and — as content explicitly required alongside the partial-fractions cases, not optional extras — $\frac{x}{x^2+5}$ and $\frac{2}{(2x-1)^4}$ , with the guidance itself noting the last two are covered "(see P3 section 5.2)". The first two are direct instances of the two general forms just derived: $\int\frac{2}{3x+5}\,dx=\frac{2}{3}\ln|3x+5|+c$ (ln-form, $n=1$ ), and $\int\frac{3}{(x-1)^2}\,dx=-\dfrac{3}{x-1}+c$ (power-form, $n=2$ , no logarithm anywhere in the answer). $\int\frac{2}{(2x-1)^4}\,dx=-\dfrac{1}{3(2x-1)^3}+c$ is the same power-form pattern at $n=4$ — the general formula derived above applied directly to the spec's own highest-power named example.

$\frac{x}{x^2+5}$ is genuinely different from every other example in this lesson: $x^2+5$ is an irreducible quadratic, and spec 2.1 explicitly excludes quadratic factors like $(x^2+a)$ from what this course decomposes — this integrand was never meant to be split into partial fractions at all. It integrates instead (the prerequisite lesson's own technique, spec 6.3's own "P3 section 5.2" cross-reference): the numerator $x$ is exactly $\frac{1}{2}$ of the derivative of the denominator, $2x$ , so $\int\frac{x}{x^2+5}\,dx=\frac{1}{2}\ln(x^2+5)+c$ — no modulus needed, since $x^2+5$ is always positive. Spec 6.3 groups this alongside the partial-fractions cases in its own guidance column, but the required SKILL here is recognition, not decomposition.

Marked, line by line

Given that $\dfrac{8x^2+8x-3}{(x-2)(x+1)(2x+1)}\equiv\dfrac{3}{x-2}-\dfrac{1}{x+1}+\dfrac{4}{2x+1}$ — the decomposition established in the worked chain earlier in this lesson — find the exact value of $\displaystyle\int_3^4 \dfrac{8x^2+8x-3}{(x-2)(x+1)(2x+1)}\,dx$ . (6) VERIDIAN-original numeric scenario, built directly on this lesson's own earlier decomposition rather than a real past-paper question.

6 marks available

01
Integrate each term using $\int\frac{1}{ax+b}\,dx=\frac{1}{a}\ln|ax+b|+c$ : $\int\frac{3}{x-2}\,dx=3\ln|x-2|$ and $\int\frac{-1}{x+1}\,dx=-\ln|x+1|$ — both denominators have $x$ -coefficient 1, so no extra scaling is needed for these two terms.
Method mark for attempting the ln-form integration on at least one term with the correct general structure.
M1
02
The third term's denominator has $x$ -coefficient 2, not 1: $\int\frac{4}{2x+1}\,dx=4\times\frac{1}{2}\ln|2x+1|=2\ln|2x+1|$ — the $\frac{1}{a}=\frac{1}{2}$ scaling factor is applied first, THEN multiplied by the numerator's own coefficient of 4.
Method mark specifically for correctly identifying and applying the 1/a=1/2 scaling on the 2x+1 term — the exact step this question is built to test.
M1
03
Indefinite integral: $3\ln|x-2|-\ln|x+1|+2\ln|2x+1|+c$ .
Accuracy mark for the fully correct antiderivative, combining all three terms.
A1
04
At $x=4$ : $3\ln 2-\ln 5+2\ln 9$ . At $x=3$ : $3\ln 1-\ln 4+2\ln 7=-\ln 4+2\ln 7$ .
Method mark for substituting both limits correctly into the antiderivative.
M1
05
Definite integral $=(3\ln 2-\ln 5+2\ln 9)-(-\ln 4+2\ln 7)=\ln\dfrac{2^3\times 9^2\times 4}{5\times 7^2}=\ln\dfrac{2592}{245}$ .
Dependent method mark for combining the evaluated limits into a single exact logarithm, dependent on both limits above being substituted correctly.
dM1
06
Exact value: $\ln\dfrac{2592}{245}$ ( $\approx 2.36$ to 3 s.f.).
Accuracy mark for the exact value in lowest terms (2592 and 245 share no common factor). An exact log form is required here, not a rounded decimal, on the same general-principles policy the facts bank documents for this paper.
A1

Worked, in full

Integrate $\dfrac{2}{3x+1}+\dfrac{3}{x-2}-\dfrac{1}{(x-2)^2}$ — the method-comparison's own decomposition, one ln term and one power term side by side

01
$\displaystyle\int\frac{2}{3x+1}\,dx=2\times\frac{1}{3}\ln|3x+1|=\frac{2}{3}\ln|3x+1|$ — ln-form, since the power is $-1$ ; the $\frac{1}{a}=\frac{1}{3}$ scaling factor is applied before multiplying by the numerator's coefficient of 2.
Earns: M1 — correct ln-form integration with the 1/a=1/3 scaling factor correctly applied.
02
$\displaystyle\int\frac{3}{x-2}\,dx=3\ln|x-2|$ — also ln-form, but this denominator's own $x$ -coefficient is 1, so no extra scaling is needed.
Earns: A1 — correct ln-form integration; recognising when the 1/a factor is trivially 1 is as much a real check as applying it when it isn't.
03
$\displaystyle\int\frac{-1}{(x-2)^2}\,dx$ — this is a REPEATED linear factor, power $-2$ , not $-1$ : no logarithm applies here at all. Using $\int\frac{1}{(ax+b)^n}\,dx=\frac{-1}{(n-1)a(ax+b)^{n-1}}+c$ with $a=1$ , $n=2$ : $\int\frac{1}{(x-2)^2}\,dx=\frac{-1}{x-2}$ , so $\int\frac{-1}{(x-2)^2}\,dx=\frac{1}{x-2}$ .
Earns: A1 — correct power-form integration, and specifically for NOT reaching for a logarithm on a squared bracket, the exact distinction this lesson's own mechanism block derives.
04
Combine: $\displaystyle\int\left(\frac{2}{3x+1}+\frac{3}{x-2}-\frac{1}{(x-2)^2}\right)dx=\frac{2}{3}\ln|3x+1|+3\ln|x-2|+\frac{1}{x-2}+c$ .
Earns: A1 — correctly combined final antiderivative, including the constant of integration.
05
Check by differentiating the last term alone: $\dfrac{d}{dx}\left[\dfrac{1}{x-2}\right]=\dfrac{-1}{(x-2)^2}$ — exactly the third piece of the original integrand. ✓
Earns: Nothing — the mark scheme has run out, but this single-term check is the fastest way to catch a sign error on the power-rule term specifically.

Spec 4.1's other application of the same decomposition: expansion in ascending powers of x

Spec 4.1's own guidance — quoted in full in WMA14-verified-facts.md §1 — asks for two things: "students should be able to obtain the expansion of (ax + b)ⁿ, and the expansion of rational functions by decomposition into partial fractions." The prerequisite lesson covers the first half directly; this is where the second half lives — the SAME decomposition already found above for the method-comparison question, $\dfrac{11x^2-26x+1}{(3x+1)(x-2)^2}\equiv\dfrac{2}{3x+1}+\dfrac{3}{x-2}-\dfrac{1}{(x-2)^2}$ , expanded in ascending powers of x instead of integrated. Nothing about the decomposition changes — only what happens to it next.

Each single-linear-factor term still needs the same 'factor out a constant' move the prerequisite lesson uses to reach $(1+u)^n$ before the general series applies: $\dfrac{2}{3x+1}=2(1+3x)^{-1}$ is already in that shape with $u=3x$ , but $\dfrac{3}{x-2}$ is not — $x-2=-2\left(1-\frac12x\right)$ , so $\dfrac{3}{x-2}=-\dfrac32\left(1-\frac12x\right)^{-1}$ , with $u=-\frac12x$ . The repeated-factor term needs the same rewriting, squared: $-\dfrac{1}{(x-2)^2}=-\dfrac{1}{4\left(1-\frac12x\right)^2}=-\dfrac14\left(1-\frac12x\right)^{-2}$ — same $u$ as the term above it, since it's the same bracket, but $n=-2$ instead of $n=-1$ .

One genuinely new point, not needed when a single bracket is expanded on its own: this expansion is a SUM of three separate series, and the sum only equals the original function where every one of the three converges at once. $2(1+3x)^{-1}$ needs $|3x|<1$ , i.e. $|x|<\frac13$ ; both $\left(1-\frac12x\right)^{-1}$ and $\left(1-\frac12x\right)^{-2}$ need $|x|<2$ . The combined range is the MOST RESTRICTIVE of the three — $|x|<\frac13$ — not the widest and not an average: outside $|x|<\frac13$ the first series has already stopped converging, however well-behaved the other two remain. This is genuine reasoning about what a sum of series requires, not a sourced examiner-report finding — no quote in this unit's research bank documents students being tested on this specific point — flagged honestly on the same labelling discipline this lesson's own trap-taxonomy already applies to its two unsourced entries.

Worked, in full

Expand $\dfrac{2}{3x+1}+\dfrac{3}{x-2}-\dfrac{1}{(x-2)^2}$ in ascending powers of x, up to and including the term in $x^2$ , and state the range of validity — the method-comparison's own decomposition, spec 4.1's combined technique applied to it. VERIDIAN-original, not a reproduction of any past-paper question.

01
Rewrite every term as $k(1+u)^n$ . $\dfrac{2}{3x+1}=2(1+3x)^{-1}$ needs no extra factoring — its constant term is already 1, so $u=3x$ , $n=-1$ . For $\dfrac{3}{x-2}$ : factor $-2$ out of $x-2$ : $x-2=-2\left(1-\frac12x\right)$ , so $\dfrac{3}{x-2}=\dfrac{3}{-2\left(1-\frac12x\right)}=-\dfrac32\left(1-\frac12x\right)^{-1}$ , $u=-\frac12x$ , $n=-1$ . And $-\dfrac{1}{(x-2)^2}=-\dfrac{1}{\left[-2\left(1-\frac12x\right)\right]^2}=-\dfrac{1}{4\left(1-\frac12x\right)^2}=-\dfrac14\left(1-\frac12x\right)^{-2}$ , the same $u=-\frac12x$ as the term above it, but $n=-2$ .
Earns: M1 — attempts to write every term as k(1+u)ⁿ by factoring a constant out of each linear bracket, including the repeated-factor term.
02
Expand each bracket to the $x^2$ term using the general series $(1+u)^n=1+nu+\frac{n(n-1)}{2!}u^2+\ldots$ : $2(1+3x)^{-1}=2\left[1-3x+9x^2-\ldots\right]=2-6x+18x^2-\ldots$
Earns: A1 — correct expansion of the first, already-(1+u)-shaped term.
03
$-\dfrac32\left(1-\frac12x\right)^{-1}=-\dfrac32\left[1+\frac12x+\frac14x^2+\ldots\right]=-\dfrac32-\dfrac34x-\dfrac38x^2-\ldots$
Earns: A1 — correct expansion of the second term, n=−1 applied to u=−½x.
04
$-\dfrac14\left(1-\frac12x\right)^{-2}=-\dfrac14\left[1+x+\dfrac34x^2+\ldots\right]=-\dfrac14-\dfrac14x-\dfrac{3}{16}x^2-\ldots$ — genuinely different coefficients from stage 3's, despite sharing the same u: the repeated factor's own n=−2 changes every coefficient the general formula produces, not just the power on the bracket.
Earns: A1 — correct expansion of the repeated-factor term, using n=−2 rather than reusing stage 3's n=−1 coefficients.
05
Add all three, term by term: constant $2-\frac32-\frac14=\frac14$ ; $x$ -term $-6-\frac34-\frac14=-7$ ; $x^2$ -term $18-\frac38-\frac{3}{16}=\frac{279}{16}$ . So $\dfrac{2}{3x+1}+\dfrac{3}{x-2}-\dfrac{1}{(x-2)^2}=\dfrac14-7x+\dfrac{279}{16}x^2-\ldots$
Earns: A1 — correctly combined final expansion.
06
State the range: the three individual conditions found in stage 1 are $|x|<\frac13$ , $|x|<2$ and $|x|<2$ — the combined series only equals the original function where ALL of them hold, so the range is the smallest one: $|x|<\frac13$ .
Earns: B1 — correct range, taken as the most restrictive of the three individual conditions, not the widest.
07
Check at $x=0.02$ (well inside $|x|<\frac13$ ): original function $=\dfrac{2}{1.06}+\dfrac{3}{-1.98}-\dfrac{1}{(-1.98)^2}\approx0.116565$ ; three-term expansion $=\frac14-7(0.02)+\frac{279}{16}(0.02)^2\approx0.116975$ — agreeing to 3 significant figures. ✓
Earns: Nothing — not a markable step, but the fastest way to catch an arithmetic slip in any one of the three separate expansions before it's submitted.

Complete it yourself

Complete the chain — decompose and then integrate $\dfrac{3x+2}{(x-1)(x+4)}$

01
Set up the identity with one unknown per distinct linear factor: $\dfrac{3x+2}{(x-1)(x+4)}\equiv\dfrac{A}{x-1}+\dfrac{B}{x+4}$ . Clear denominators: $3x+2\equiv A(x+4)+B(x-1)$ .
02
Substitute $x=1$ (the root of $x-1$ ), which vanishes the $B$ term: $3(1)+2=A(5)\Rightarrow 5=5A\Rightarrow A=1$ .

In your own words

In one sentence: why does a repeated linear factor $(cx+d)^2$ in the denominator produce ONE logarithm term and ONE non-logarithm (power-rule) term when integrated, rather than two logarithms?

Named traps

extra-power-created-by-over-multiplying-a-repeated-factor: A real examiner report on a repeated-linear-factor decomposition (a DIFFERENT question from this lesson's own (3x+1)(x−2)² example, with its own denominator built around a bracket of (2x+1)) records: "Relatively few students made the error of multiplying by the product of the three denominators so having (2x+1)³" (Jan 2024, Q2(a)) — i.e. treating the repeated bracket as though it needed multiplying by itself an extra, unneeded time when clearing denominators, turning a correct squared bracket into a wrong cubed one. Real, but a genuine minority error ("relatively few"), not the dominant failure mode on this question type — that distinction is worth keeping honest rather than inflating.
simultaneous-equations-more-error-prone-than-substitution: A real examiner report on the same question records a genuine, quotable method-choice recommendation: simultaneous equations built from expanded coefficients, used by a significant number of candidates, were "more prone to error than attempts via substitution" (Jan 2024, Q2(a)) — exactly the finding this lesson's own method-comparison block anchors on. Both methods are legitimate and both reach full marks; the real record's point is which one real scripts actually get wrong more often, not that one method is invalid.
sign-lost-in-a-long-division-remainder: On a question comparing the identity method against long division as two legitimate routes to the same decomposition: "The most common error with candidates who used this method was to set the numerator of the partial fraction equal to 6 rather than -6 and then forgetting the negative when the values were put back into the expression" (Oct 2021, Q3(a)) — a sign lost while forming the remainder during division, then lost a SECOND time when substituting values back in. This lesson's own marked-solution above recreates the same failure class with its own numbers (2x+7 corrupted to 2x−1), not the real question's own 6/−6.
1-over-a-scaling-factor-omitted-on-a-ln-integral: The single most robustly documented error in this whole topic — confirmed independently in TWO separate series: "a common error was to omit the ½ in ∫1/(2x−1) dx = ½ ln(2x−1)" (Oct 2022, Q2), and again, on a different question testing the same skill: "A number of students failed to divide the second logarithm by two" (Jan 2024, Q2(b)). Both quotes describe the identical omission — the reciprocal of the x-coefficient inside the bracket, dropped from the front of the logarithm — this lesson's own marked-solution above is built specifically to anchor on this two-series-confirmed finding.
ln-rule-misapplied-to-a-repeated-factor: Flagged honestly as VERIDIAN-original pedagogical inference, NOT a sourced examiner-report finding — no quote in this unit's research bank documents this specific confusion. The two integrand shapes $\frac{1}{ax+b}$ and $\frac{1}{(ax+b)^2}$ look almost identical on the page, and it would be a natural mistake to apply the ln rule to BOTH, writing something like $\ln|(x-2)^2|$ for the squared case instead of the correct power-rule answer $\frac{-1}{x-2}$ . Included here because the two integrand shapes really are this visually close, and this lesson's own mechanism block above exists specifically to derive why they need genuinely different rules — not because a real report has been found documenting students making exactly this error on this paper.
degree-condition-overlooked-before-attempting-the-basic-form: Anchored to spec 2.1's own explicit clause — "The degree of the numerator may equal or exceed the degree of the denominator" — rather than to a specific examiner-report quote describing students missing this: no citation in this unit's research bank documents this exact failure in these terms. Attempting $\frac{A}{x-a}+\frac{B}{x-b}$ directly on a fraction whose numerator's degree already matches the denominator's leads nowhere, because no choice of A and B can make that form reproduce a fraction that doesn't tend to 0 as x grows large. Worth checking degree BEFORE choosing a decomposition form, not after the first attempt fails.

Retrieval — with feedback on every choice

Question 1

2 marks

Using the cover-up rule on $\dfrac{5x-5}{(x-3)(x+2)}\equiv\dfrac{A}{x-3}+\dfrac{B}{x+2}$ , what is the value of A?

Question 2

2 marks

$\dfrac{4x^2-1}{(2x+3)(x-1)}$ — before this fraction can be decomposed with $\dfrac{A}{2x+3}+\dfrac{B}{x-1}$ , what needs to happen first?

Question 3

2 marks

$\displaystyle\int \frac{7}{4x-3}\,dx=$ ?

Question 4

2 marks

$\displaystyle\int \frac{5}{(x+2)^2}\,dx=$ ?

Question 5

2 marks

A denominator is $(x+1)(2x-3)^2$ . How many unknown constants does its full partial-fraction decomposition need, and why?

Question 6

2 marks

$\dfrac{11x^2-26x+1}{(3x+1)(x-2)^2}$ is decomposed and each term expanded separately, giving individual ranges of validity $|x|<\dfrac13$ , $|x|<2$ and $|x|<2$ . What is the range of validity of the COMBINED expansion (all three series added together)?

Reference — not a study method, a lookup

Distinct linear factors (ax+b)(cx+d)(ex+f): one term each, found by cover-up (substitute each factor's own root).
Repeated factor (cx+d)²: TWO terms — B/(cx+d) + C/(cx+d)² — never one. A denominator of degree n needs n unknowns total.
Numerator degree ≥ denominator degree: divide first, or use the identity method with an added constant — A/(x-a)+B/(x-b) alone can't represent it.
∫1/(ax+b)dx = (1/a)ln|ax+b| + c. The 1/a is real and load-bearing — confirmed the single most-documented slip in this topic, in TWO series.
∫1/(ax+b)ⁿdx for n≥2 is POWER-form, no logarithm at all: -1/[(n-1)a(ax+b)ⁿ⁻¹] + c.
x/(x²+5)-style integrands are NOT partial fractions (quadratic factors excluded by spec 2.1) — they integrate by recognition instead.
Expanding a decomposition in x (spec 4.1): factor a constant out of EVERY term to reach k(1+u)ⁿ, same as for a single bracket. The combined range of validity is the SMALLEST of the individual |u|<1 ranges, not the largest.

Not affiliated with or endorsed by Pearson Edexcel. Every quotation attributed to an examiner report in this lesson was checked against WMA14-verified-facts.md §5.2, using only text genuinely inside quotation marks in that document, not the facts-bank compiler's own paraphrased framing around those quotes — concretely, "more prone to error than attempts via substitution" (Jan 2024, Q2(a)), "Relatively few students made the error of multiplying by the product of the three denominators so having (2x+1)³" (Jan 2024, Q2(a), a DIFFERENT denominator from this lesson's own worked examples), "The most common error with candidates who used this method was to set the numerator of the partial fraction equal to 6 rather than -6 and then forgetting the negative when the values were put back into the expression" (Oct 2021, Q3(a)), "a common error was to omit the ½ in ∫1/(2x−1) dx = ½ ln(2x−1)" (Oct 2022, Q2), and "A number of students failed to divide the second logarithm by two" (Jan 2024, Q2(b)) — the last two are the two-series-confirmed 1/a-omission finding this lesson's marked-solution above is built directly around. Two items in the trap-taxonomy above are explicitly flagged as NOT sourced from an examiner report — the ln-rule-vs-power-rule confusion and the degree-condition omission — because no quote in this unit's narrower research bank (10 series with at least one document, 4 full examiner reports — see the facts bank's own §6 and §8) documents either failure in these specific terms; both are labelled as VERIDIAN pedagogical inference, anchored to real mechanism or spec-clause reasoning rather than to a fabricated citation. Every worked example, prequestion, MCQ, marked-solution, method-comparison and chain-drill question in this lesson — including every numerator, every denominator, every mark tariff, and both deliberately-wrong scripts — is VERIDIAN-original, checked by independent symbolic and numeric differentiation before being written in; none reproduces a real Pearson question's own numbers, wording, or denominator. The four spec-named integrands used directly (2/(3x+5), 3/(x−1)², x/(x²+5), 2/(2x−1)⁴) come from spec 2.1 and 6.3's own guidance text, verified verbatim in WMA14-verified-facts.md §1, not from any past paper. One further block, added 2026-09-10 and tagged SPEC_4_1: it expands the method-comparison's own decomposition in ascending powers of x instead of integrating it, closing a gap an adversarial review found in spec 4.1's own guidance ("the expansion of rational functions by decomposition into partial fractions", named verbatim in WMA14-verified-facts.md §1) that neither this lesson nor binomial-expansion-for-rational-n.ts previously covered. Every coefficient in it was checked with sympy's own series() before being written in, the same discipline as everywhere else in this file; nothing in it — including the range-of-validity-as-intersection reasoning — is drawn from an examiner report, and the block's own text says so.

Question 12 marks

Using the cover-up rule on $\dfrac{5x-5}{(x-3)(x+2)}\equiv\dfrac{A}{x-3}+\dfrac{B}{x+2}$ , what is the value of A?

$A=2$
Correct. Substituting $x=3$ (covering up the $(x-3)$ factor): $A=\dfrac{5(3)-5}{3+2}=\dfrac{10}{5}=2$ .
B $A=\dfrac{5(-2)-5}{-2-3}$ , i.e. using $x=-2$
This finds B, not A — $x=-2$ is the root of the OTHER factor, $(x+2)$ , and covering up that bracket isolates B. Finding A requires substituting the root of the bracket A's OWN denominator doesn't contain, which is $x=3$ .
C $A=\dfrac{5(3)-5}{3-2}=10$
The denominator used here is wrong — after covering up $(x-3)$ , what remains is $(x+2)$ evaluated at $x=3$ , which is $3+2=5$ , not $3-2$ . The sign inside the remaining bracket has to match the bracket as written, not be re-derived from the substituted value alone.
DThe cover-up rule cannot find A directly — simultaneous equations are required
The cover-up rule works for every distinct linear factor's own coefficient, one substitution each — that's exactly what makes it faster than simultaneous equations for this shape of denominator, not a method that only works for some of the unknowns.

Traps tested: Cover up root matched to wrong unknown · Remaining bracket evaluated with wrong sign · Cover up rule scope understated

Question 22 marks

$\dfrac{4x^2-1}{(2x+3)(x-1)}$ — before this fraction can be decomposed with $\dfrac{A}{2x+3}+\dfrac{B}{x-1}$ , what needs to happen first?

A polynomial part needs to be extracted first, since the numerator's degree (2) is not smaller than the denominator's degree (2)
Correct — the same degree condition from this lesson's own worked example: numerator and denominator degrees are equal here, so $\frac{A}{2x+3}+\frac{B}{x-1}$ alone can never represent this fraction, whatever A and B turn out to be.
BNothing — this form is already correct, since the denominator only has two distinct linear factors
The number of factors in the denominator isn't what the degree condition is about — it's the numerator's own degree relative to the denominator's total degree, and here they're equal (both 2).
CThe fraction needs to be simplified by cancelling a common factor first
There's no shared factor between $4x^2-1$ and $(2x+3)(x-1)$ to cancel — $4x^2-1$ factorises as $(2x-1)(2x+1)$ , neither of which matches either denominator factor. The issue is purely the numerator's degree, not a hidden common factor.
DAn extra term with denominator $(2x+3)(x-1)$ needs to be added to the decomposition
Adding a term whose denominator is the full original product would just undo the whole point of decomposing — every genuine partial-fraction term has a denominator that's a single factor (or a power of one), never the original product again.

Traps tested: Degree condition overlooked · Wrong fix applied cancellation instead of division · Undecomposed product reintroduced as a fix

Question 32 marks

$\displaystyle\int \frac{7}{4x-3}\,dx=$ ?

$\dfrac{7}{4}\ln|4x-3|+c$
Correct. The $x$ -coefficient inside the bracket is 4, so the $\frac{1}{4}$ scaling factor applies, multiplied by the numerator's own coefficient of 7.
B $7\ln|4x-3|+c$
This is the two-series-confirmed omission this lesson's own marked-solution is built around — the reciprocal-of-coefficient scaling factor (here, $\frac{1}{4}$ ) has been dropped entirely.
C $\dfrac{7}{4}\ln|4x-3|^4+c$
Differentiating this gives $\frac{7}{4}\times\frac{4(4x-3)^3\times 4}{(4x-3)^4}=\frac{28}{4x-3}$ — four times too large. The power 4 belongs nowhere in this ln-form answer; that shape is for a REPEATED linear factor in the original integrand, which $\frac{7}{4x-3}$ is not.
D $-\dfrac{7}{4(4x-3)}+c$
This is the antiderivative for $\frac{7}{(4x-3)^2}$ (power-form), not $\frac{7}{4x-3}$ (ln-form) — differentiate it and you get $\frac{7}{(4x-3)^2}$ , not the required integrand.

Traps tested: Reciprocal of coefficient scaling factor omitted · Power form and ln form conflated · Power form applied where ln form is needed

Question 42 marks

$\displaystyle\int \frac{5}{(x+2)^2}\,dx=$ ?

$-\dfrac{5}{x+2}+c$
Correct — power-form, since the bracket is squared: reversing the chain rule on $(x+2)^{-1}$ gives $-\frac{1}{x+2}$ , multiplied by the numerator's coefficient of 5.
B $5\ln|x+2|^2+c$
This applies the ln rule to a REPEATED factor, exactly the trap this lesson's own worked chain is built to show the opposite of. Differentiating this gives $\frac{10}{x+2}$ , nothing like the required integrand $\frac{5}{(x+2)^2}$ .
C $5\ln|x+2|+c$
This is the antiderivative of $\frac{5}{x+2}$ (power $-1$ ), not $\frac{5}{(x+2)^2}$ (power $-2$ ) — a repeated linear factor never integrates to a plain logarithm.
D $\dfrac{5}{2(x+2)}+c$
The sign is wrong — reversing the chain rule on $(x+2)^{-1}$ gives $-\frac{1}{x+2}$ , not $+\frac{1}{x+2}$ , and there's no factor of 2 in this denominator's coefficient to justify the $\frac{1}{2}$ either.

Traps tested: Ln rule misapplied to a repeated factor · Sign and coefficient both lost in power rule reversal

Question 52 marks

A denominator is $(x+1)(2x-3)^2$ . How many unknown constants does its full partial-fraction decomposition need, and why?

Three — one for $(x+1)$ , and two for $(2x-3)^2$ (one term at the single power, one at the squared power)
Correct — this lesson's own mechanism block derives exactly why: a proper numerator over this degree-3 denominator can have up to 3 independent coefficients, and the right-hand side needs that many independent unknowns to be able to match any of them.
BTwo — one for $(x+1)$ , one for $(2x-3)^2$ as a single term
A single term for the repeated factor gives only 2 unknowns total, which is enough to match a numerator with only 2 independent coefficients — never a general proper numerator over a degree-3 denominator, which needs 3.
CFour — one for $(x+1)$ , and three for $(2x-3)^2$ , since it's squared
The repeated factor needs exactly as many terms as its own power, not one more — $(2x-3)^2$ needs terms at powers 1 and 2, which is two terms, not three.
DIt depends on the numerator — a smaller numerator needs fewer unknowns
The count is fixed by the DENOMINATOR's shape alone, so that the form can represent ANY proper numerator over it — it doesn't shrink for a particular, simpler-looking numerator that happens to be given.

Traps tested: Repeated factor single power term omitted · Repeated factor term count overstated · Term count treated as numerator dependent

Question 62 marks

$\dfrac{11x^2-26x+1}{(3x+1)(x-2)^2}$ is decomposed and each term expanded separately, giving individual ranges of validity $|x|<\dfrac13$ , $|x|<2$ and $|x|<2$ . What is the range of validity of the COMBINED expansion (all three series added together)?

$|x|<\dfrac13$ — the smallest of the three, since the sum only equals the original function where every one of its three series converges at once.
Correct. Outside $|x|<\frac13$ the $2(1+3x)^{-1}$ term's own series has already stopped converging, however well-behaved the other two terms remain — a sum of series is only as valid as its LEAST valid piece.
B $|x|<2$ — the widest of the three, since two of the three terms are valid there.
Backwards. A majority vote among the three ranges isn't what determines validity — the sum needs EVERY term to converge, so the widest range is exactly the one the third term's own $|x|<\frac13$ condition rules out.
C $\dfrac13<|x|<2$ — an interval starting where the first series stops and ending where the other two do.
This treats the three ranges as though they combined into a band between two bounds, rather than each being its own condition measured from $x=0$ . At, say, $x=0.1$ (which lies OUTSIDE this proposed interval but inside $|x|<\frac13$ ), all three individual series in fact converge — this interval excludes values that are genuinely valid.
DIt can't be stated without first finding the exact $x^2$ coefficient of the combined expansion.
The range depends only on where each individual bracket's own series converges — which is fixed once $u$ is identified for each term (stage 1 of the worked chain above) — not on how far the expansion has been carried out or how accurate any one coefficient is.

Traps tested: Overall range taken as widest not narrowest · Individual ranges combined as an interval not an intersection from zero · Range conflated with coefficient accuracy

Practice this for real

This site teaches the mechanism; the exam is sat on Pearson's own real questions. Go find and attempt these yourself — nothing here substitutes for actually sitting a timed paper.

Pearson's official past-papers portal

Select International Advanced Level → Mathematics → any series, then look for WMA14.

Partial Fractions — Decomposition and Integration

What partial-fraction decomposition is, and why this paper pairs it with integration

Why a partial-fraction identity can be checked at chosen values of x — and why matching coefficients works too

Decompose 8x2+8x−3(x−2)(x+1)(2x+1)\dfrac{8x^2+8x-3}{(x-2)(x+1)(2x+1)}(x−2)(x+1)(2x+1)8x2+8x−3​ — three distinct linear factors, spec 2.1's own named shape (ax+b)(cx+d)(ex+f)(ax+b)(cx+d)(ex+f)(ax+b)(cx+d)(ex+f)

Why a repeated factor needs two terms, not one

When the numerator's degree isn't smaller than the denominator's

Where the 1/a in ∫1/(ax+b)dx = (1/a)ln|ax+b| + c actually comes from

The rest of spec 6.3's own named guidance

Integrate 23x+1+3x−2−1(x−2)2\dfrac{2}{3x+1}+\dfrac{3}{x-2}-\dfrac{1}{(x-2)^2}3x+12​+x−23​−(x−2)21​ — the method-comparison's own decomposition, one ln term and one power term side by side

Spec 4.1's other application of the same decomposition: expansion in ascending powers of x

Complete the chain — decompose and then integrate 3x+2(x−1)(x+4)\dfrac{3x+2}{(x-1)(x+4)}(x−1)(x+4)3x+2​

Deep dive

Practice this for real

Decompose $\dfrac{8x^2+8x-3}{(x-2)(x+1)(2x+1)}$ — three distinct linear factors, spec 2.1's own named shape $(ax+b)(cx+d)(ex+f)$

Integrate $\dfrac{2}{3x+1}+\dfrac{3}{x-2}-\dfrac{1}{(x-2)^2}$ — the method-comparison's own decomposition, one ln term and one power term side by side

Complete the chain — decompose and then integrate $\dfrac{3x+2}{(x-1)(x+4)}$