Paper 3 only. Reverse the process of adding fractions: break a rational function into simple pieces, ready for binomial expansion and integration. Three denominator types are on the syllabus.
Choose the form from the denominator, multiply through to get an identity in \(x\), then find the constants by substituting the root of each linear factor and comparing coefficients for whatever is left. A repeated factor \((cx + d)^2\) needs BOTH \(\dfrac{B}{cx + d}\) and \(\dfrac{C}{(cx + d)^2}\); a quadratic factor \(cx^2 + d\) needs a numerator \(Bx + C\). If the numerator has the same degree as the denominator, include a constant term as well.
Section 01
Choose the form first.
Everything depends on writing down the correct form before you calculate anything. Factorise the denominator fully, then match it to one of the three syllabus types. This form assumes the numerator has a lower degree than the denominator; when the degrees are EQUAL, add a constant term too (last section). A numerator of higher degree is not on the syllabus.
Factorise the denominator before choosing the form. \(x^2 - 2x - 3\) is not a 'quadratic factor' needing \(Bx + C\): it is \((x - 3)(x + 1)\). Only a quadratic that does not factorise, such as \(x^2 + 3\), gets the \(Bx + C\) numerator.
Section 02
Three linear factors.
With three different linear factors there are three constants, and each one comes from substituting the value of \(x\) that makes its own factor zero. Fractions such as \(x = -\dfrac{1}{2}\) are normal; do the arithmetic carefully.
Worked example Three distinct factors, one non-unit
When substituting \(x = -\dfrac{1}{2}\), every bracket must be evaluated at \(-\dfrac{1}{2}\): \((x - 1) = -\dfrac{3}{2}\), not \(-1\). A single slip here makes \(C\) wrong, so always check with a coefficient comparison.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
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Which is the correct form for \(\dfrac{x + 5}{(x - 1)(x + 3)^2}\)?
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Which is the correct form for \(\dfrac{1}{(x + 1)(x^2 + 4)}\)?
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\(\dfrac{7}{(x - 2)(x + 5)} \equiv \dfrac{A}{x - 2} + \dfrac{B}{x + 5}\). What is \(A\)?
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In \(8x - 1 \equiv A(2x - 1) + B(x + 1)\), which value of \(x\) gives \(B\) directly?
Section 03
A repeated factor.
For \((cx + d)^2\) you need two fractions, \(\dfrac{B}{cx + d}\) and \(\dfrac{C}{(cx + d)^2}\). Substitution finds \(A\) and \(C\) quickly; \(B\) usually comes from comparing the \(x^2\) coefficients.
Writing only \(\dfrac{A}{x + 1} + \dfrac{C}{(2x + 1)^2}\) loses the middle term. With two unknowns the identity cannot hold for every \(x\), and you will find the 'answers' disagree when you check.
Section 04
An irreducible quadratic factor.
When the denominator contains \(cx^2 + d\) (with \(c, d > 0\), so it has no real roots), its fraction needs a linear numerator \(Bx + C\). Only one substitution kills a term; compare coefficients for the rest.
\(\dfrac{B}{x^2 + 3}\) with a constant numerator is the wrong form: it gives too few unknowns, and the \(x\) coefficients will not match. Over an irreducible quadratic always write \(Bx + C\).
Section 05
Numerator of equal degree: add a constant.
When the numerator has the SAME degree as the denominator, the fraction is improper: it contains a whole-number part. Either divide first and split the remainder, or simply include a constant term \(A\) in the form and find it by comparing the highest powers.
Worked example Equal degrees
Express \(\dfrac{2x^2 + x + 3}{(x - 1)(2x + 1)}\) in partial fractions.
The denominator expands to \(2x^2 - x - 1\), also degree 2, so the form is \(A + \dfrac{B}{x - 1} + \dfrac{C}{2x + 1}\).
Divide first: \(2x^2 + x + 3 = 1 \times (2x^2 - x - 1) + (2x + 4)\), so \(A = 1\) and the fraction is \(1 + \dfrac{2x + 4}{(x - 1)(2x + 1)}\).
Leaving out the constant term is the classic error here. Without it the identity \(2x^2 + x + 3 \equiv B(2x + 1) + C(x - 1)\) is impossible: the right side has no \(x^2\) term. Compare the degrees of the numerator and denominator before you write down the form.
Stage 2 · Exam
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