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LearnYMA01WMA14 · Pure Mathematics 4IntegrationIntegration by parts
Integration · Subtopic 3 of 7

Integration by Parts

The reverse of the product rule: ∫udvdx dx=uv−∫vdudx dx\displaystyle\int u\frac{dv}{dx}\,dx=uv-\int v\frac{du}{dx}\,dx. How to choose uu, when to use it twice, and the trick for ∫ln⁡x dx\int\ln x\,dx.

Cambridge 9709 Paper 3Edexcel YMA01 WMA14Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1AQA 7357 Paper 2≈ 50 minRevision 6 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Derive integration by parts from the product rule.
  • Choose uu so that dudx\dfrac{du}{dx} is simpler than uu.
  • Integrate ln⁡x\ln x and xnln⁡xx^n\ln x by taking u=ln⁡xu=\ln x.
  • Use parts twice when needed, and evaluate definite integrals exactly.
Pause and think

Differentiate xexxe^x with the product rule. Rearrange the result to find ∫xex dx\displaystyle\int xe^x\,dx.

Section 2 of 14

Core idea

Start hereUndoing the product rule

Differentiating a product is easy (product rule). Integrating one, like ∫xex dx\int xe^x\,dx, isn't: no inner function to substitute. Integration by parts reverses the product rule, trading a hard integral for (hopefully) an easier one.

The big idea

∫udvdxdx=uv−∫vdudxdx\displaystyle\int u\frac{dv}{dx}dx=uv-\int v\frac{du}{dx}dx. Choose uu to get simpler when differentiated (LATE: Logs, Algebra, Trig, Exponentials).

1. Derivation

From the product rule
  1. ddx(uv)=udvdx+vdudx\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}
    Product rule.
  2. uv=∫udvdxdx+∫vdudxdxuv=\int u\frac{dv}{dx}dx+\int v\frac{du}{dx}dx
    Integrate both sides.
  3. ∫udvdxdx=uv−∫vdudxdx\int u\frac{dv}{dx}dx=uv-\int v\frac{du}{dx}dx
    Rearrange.
Area under y = x eˣ from 0 to 1 is exactly 1
−0.50.51−0.50.511.522.533.50xyy = x eˣarea 1
[ex(x−1)]01=0−(−1)=1\big[e^x(x-1)\big]_0^1=0-(-1)=1.
Worked example∫01xex dx\int_0^1xe^x\,dx

Evaluate.

Goal
Number.
Strategy
u=xu=x (simplifies to 11), v′=exv'=e^x.
  1. u=x, u′=1;v′=ex, v=exu=x,\ u'=1;\quad v'=e^x,\ v=e^x
    Choose.
  2. [xex]01−∫01exdx[xe^x]_0^1-\int_0^1e^xdx
    Formula.
  3. =e−[ex]01=e−(e−1)=1=e-[e^x]_0^1=e-(e-1)=1
    Evaluate.
Answer11.
✓ Checkddx(xex−ex)=xex\frac{d}{dx}(xe^x-e^x)=xe^x ✓.
Worked example∫ln⁡x dx\int\ln x\,dx

Integrate.

Goal
Antiderivative.
Strategy
Write as ln⁡x⋅1\ln x\cdot1; u=ln⁡xu=\ln x.
  1. u=ln⁡x, u′=1x;v′=1, v=xu=\ln x,\ u'=\tfrac1x;\quad v'=1,\ v=x
    Choose.
  2. xln⁡x−∫x⋅1x dx=xln⁡x−x+cx\ln x-\int x\cdot\tfrac1x\,dx=x\ln x-x+c
    Formula.
Answerxln⁡x−x+cx\ln x-x+c.
✓ CheckDerivative: ln⁡x+1−1=ln⁡x\ln x+1-1=\ln x ✓.
Worked exampleTwice: ∫x2ex dx\int x^2e^x\,dx

Integrate.

Goal
Antiderivative.
Strategy
By parts twice.
  1. x2ex−∫2xexdxx^2e^x-\int2xe^xdx
    u=x2u=x^2.
  2. ∫2xexdx=2xex−2ex\int2xe^xdx=2xe^x-2e^x
    By parts again.
  3. =ex(x2−2x+2)+c=e^x(x^2-2x+2)+c
    Combine.
Answerex(x2−2x+2)+ce^x(x^2-2x+2)+c.
✓ CheckDerivative: ex(x2−2x+2+2x−2)=x2exe^x(x^2-2x+2+2x-2)=x^2e^x ✓.
Common misconceptionWrong choice of u ∫xex\int xe^x with u=exu=e^x, v′=xv'=x gives x22ex−∫x22ex\frac{x^2}2e^x-\int\frac{x^2}2e^x: worse. Use LATE.
Pause and think

∫1exln⁡x dx\int_1^ex\ln x\,dx?

Summary

  • ∫uv′=uv−∫vu′\int uv'=uv-\int vu'.
  • LATE.
  • ln⁡x\ln x: use v′=1v'=1.
Key words
By parts
Reverse product rule integration.
HistoryTaylor’s formula. The formula for integration by parts is usually credited to Brook Taylor, who published it in 1715, in the same book where his famous Taylor series appears. The idea was already being used by Leibniz and the Bernoullis.
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