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LearnYMA01WMA14 · Pure Mathematics 4Algebra and functionsPartial fractions
Algebra and functions · Subtopic 1 of 1

Partial Fractions

Splitting a complicated fraction into simpler ones, which you need for integrating and for binomial expansions: the right form for each kind of denominator, and quick ways to find the constants.

Cambridge 9709 Paper 3Edexcel YMA01 WMA14Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1AQA 7357 Paper 2≈ 45 minRevision 5 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Choose the correct form of partial fractions for linear, repeated and quadratic factors.
  • Find the constants by substituting convenient values of xx (the cover-up idea).
  • Use equating coefficients when substitution is not enough.
  • Divide first when the numerator's degree is not less than the denominator's.
Pause and think

Work out 2x−1+3x+2\dfrac{2}{x-1}+\dfrac{3}{x+2} as a single fraction.

Section 2 of 14

Core idea

Start hereUn-adding fractions

You know 2x−1+3x+2=5x+1(x−1)(x+2)\frac2{x-1}+\frac3{x+2}=\frac{5x+1}{(x-1)(x+2)}. Going backwards, splitting a complicated fraction into simple ones, is surprisingly useful: simple fractions are easy to integrate and to expand as series. The reverse process is called partial fractions.

The big idea

A proper fraction with a factorised denominator splits into simpler fractions: Ax−a\frac A{x-a} for each linear factor, Bx−a+C(x−a)2\frac B{x-a}+\frac C{(x-a)^2} for a repeated factor, Bx+Cx2+c\frac{Bx+C}{x^2+c} for an irreducible quadratic. Find constants by substituting roots and comparing coefficients.

1. Distinct linear factors

Two simple graphs add up to the complicated one
−4−224−10−55100xywhole fraction2/(x − 1)3/(x + 2)
At every xx the two dashed curves add to the solid one. Each simple piece is easy to integrate.
Worked example5x+1(x−1)(x+2)\dfrac{5x+1}{(x-1)(x+2)}

Express in partial fractions.

Goal
AA, BB.
Strategy
Multiply by the denominator to get an identity; choose xx values that kill terms.
  1. 5x+1(x−1)(x+2)≡Ax−1+Bx+2\frac{5x+1}{(x-1)(x+2)}\equiv\frac A{x-1}+\frac B{x+2}
    Form.
  2. 5x+1≡A(x+2)+B(x−1)5x+1\equiv A(x+2)+B(x-1)
    Multiply both sides by (x−1)(x+2)(x-1)(x+2).
  3. x=1: 6=3A⇒A=2x=1:\ 6=3A\Rightarrow A=2
    Kills the BB term.
  4. x=−2: −9=−3B⇒B=3x=-2:\ -9=-3B\Rightarrow B=3
    Kills the AA term.
Answer2x−1+3x+2\dfrac2{x-1}+\dfrac3{x+2}.
✓ CheckRecombine: 2(x+2)+3(x−1)(x−1)(x+2)=5x+1…\frac{2(x+2)+3(x-1)}{(x-1)(x+2)}=\frac{5x+1}{\ldots} ✓.
Key takeawayWhy substituting is allowed The identity 5x+1≡A(x+2)+B(x−1)5x+1\equiv A(x+2)+B(x-1) is true for every xx (it's a polynomial identity), so we may choose any convenient xx, even x=1x=1 where the original fraction was undefined.

2. Repeated factors

Worked example3x2+2x(x−1)2\dfrac{3x^2+2}{x(x-1)^2}

Express in partial fractions.

Goal
AA, BB, CC.
Strategy
A squared factor needs two fractions.
  1. ≡Ax+Bx−1+C(x−1)2\equiv\frac Ax+\frac B{x-1}+\frac C{(x-1)^2}
    Form.
  2. 3x2+2≡A(x−1)2+Bx(x−1)+Cx3x^2+2\equiv A(x-1)^2+Bx(x-1)+Cx
    Multiply through.
  3. x=0: A=2;x=1: C=5x=0:\ A=2;\qquad x=1:\ C=5
    Substitute roots.
  4. x2: 3=A+B⇒B=1x^2:\ 3=A+B\Rightarrow B=1
    Compare x2x^2 coefficients.
Answer2x+1x−1+5(x−1)2\dfrac2x+\dfrac1{x-1}+\dfrac5{(x-1)^2}.
✓ Checkx=2x=2: original 142=7\frac{14}{2}=7; split 1+1+5=71+1+5=7 ✓.

3. Quadratic factors and improper fractions

Worked examplex+5(x+1)(x2+3)\dfrac{x+5}{(x+1)(x^2+3)}

Express in partial fractions.

Goal
Constants.
Strategy
x2+3x^2+3 doesn't factorise: numerator Bx+CBx+C.
  1. x+5≡A(x2+3)+(Bx+C)(x+1)x+5\equiv A(x^2+3)+(Bx+C)(x+1)
    Multiply through.
  2. x=−1: 4=4A⇒A=1x=-1:\ 4=4A\Rightarrow A=1
    Substitute.
  3. x2: 0=A+B⇒B=−1;const: 5=3A+C⇒C=2x^2:\ 0=A+B\Rightarrow B=-1;\quad\text{const}:\ 5=3A+C\Rightarrow C=2
    Compare coefficients.
Answer1x+1+2−xx2+3\dfrac1{x+1}+\dfrac{2-x}{x^2+3}.
✓ Checkx=0x=0: 53\frac5{3} vs 1+23=531+\frac23=\frac53 ✓.
Worked exampleImproper: x2+1x2−1\dfrac{x^2+1}{x^2-1}

Express in partial fractions.

Goal
Polynomial + proper fractions.
Strategy
Top degree ≥ bottom degree: divide first.
  1. x2+1x2−1=1+2x2−1\frac{x^2+1}{x^2-1}=1+\frac{2}{x^2-1}
    x2+1=(x2−1)+2x^2+1=(x^2-1)+2.
  2. 2(x−1)(x+1)=1x−1−1x+1\frac2{(x-1)(x+1)}=\frac1{x-1}-\frac1{x+1}
    Linear factors.
Answer1+1x−1−1x+11+\dfrac1{x-1}-\dfrac1{x+1}.
✓ Checkx=2x=2: 53\frac53 vs 1+1−13=531+1-\frac13=\frac53 ✓.
Choosing the form

Always check by recombining or substituting a value.

Pause and think

Split 4x2−4\frac{4}{x^2-4}.

Summary

  • Forms for linear, repeated, quadratic factors.
  • Identity: substitute and compare.
  • Improper: divide first.
Key words
Proper fraction
Numerator degree < denominator degree.
Identity
True for all x.
HistoryWhy it was invented. Leibniz and Johann Bernoulli developed partial fractions around 1702 precisely to integrate rational functions. Splitting 1x2−1\dfrac{1}{x^2-1} into 12(1x−1−1x+1)\dfrac{1}{2}\left(\dfrac{1}{x-1}-\dfrac{1}{x+1}\right) turns an impossible-looking integral into two easy ones.
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