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11.6PureStretch

Integration by Parts

Learn integration by parts with worked examples for products, logarithms, repeated application and definite integrals. Choose which factor to differentiate, keep track of signs, then check your answer using the product rule.

28 min Integration
Edexcel A level Maths: 11.6 Integration By Parts (Part 1)Curated by Kepler, not produced by us. Watch it before the notes below.
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What you'll be able to do

  • State the integration by parts formula
  • Choose u and dv/dx wisely
  • Apply to products like x eˣ, x ln x
  • Handle the special case ∫ln x dx
  • Apply integration by parts twice and evaluate definite integrals
1

The formula

For a product, ∫udvdx dx=uv−∫vdudx dx\displaystyle\int u\frac{dv}{dx}\,dx = uv - \int v\frac{du}{dx}\,dx. Choose uu to be the part that gets simpler when differentiated.

∫u dvdx dx=uv−∫v dudx dx\int u\,\frac{dv}{dx}\,dx = uv - \int v\,\frac{du}{dx}\,dx
Integration by parts.
Example 1Find ∫xex dx\int x e^x\,dx.
1u=xu = x, dvdx=ex\tfrac{dv}{dx} = e^x; dudx=1\tfrac{du}{dx} = 1, v=exv = e^x.
2xex−∫ex dx=xex−ex+cx e^x - \int e^x\,dx = x e^x - e^x + c.
Answer
ex(x−1)+ce^x(x - 1) + c

Tip — LATE (Logs, Algebra, Trig, Exponentials) is a useful starting point for choosing u, not a rule that guarantees a simpler integral. Check that dv/dx can be integrated and that the remaining integral is easier.

2

Integrating ln x

A neat trick: ∫ln⁡x dx\int \ln x\,dx uses parts with u=ln⁡xu = \ln x and dvdx=1\tfrac{dv}{dx} = 1, giving xln⁡x−x+cx\ln x - x + c.

3

Repeated integration by parts

A quadratic factor becomes linear after one differentiation, so the integral left over may need a second application. Keep each subtraction in brackets until you simplify.

Example 1Find ∫x2ex dx\int x^2 e^x\,dx.
1Choose u=x2u=x^2 and dvdx=ex\tfrac{dv}{dx}=e^x, so dudx=2x\tfrac{du}{dx}=2x and v=exv=e^x.
2∫x2ex dx=x2ex−2∫xex dx\int x^2 e^x\,dx=x^2e^x-2\int xe^x\,dx.
3Apply parts again to the remaining integral: ∫xex dx=xex−ex\int xe^x\,dx=xe^x-e^x.
4Substitute the whole result: x2ex−2(xex−ex)+cx^2e^x-2(xe^x-e^x)+c.
5Check with the product rule: ddx[ex(x2−2x+2)]=ex(x2−2x+2)+ex(2x−2)=x2ex\tfrac{d}{dx}[e^x(x^2-2x+2)]=e^x(x^2-2x+2)+e^x(2x-2)=x^2e^x.
Answer
ex(x2−2x+2)+ce^x(x^2-2x+2)+c

Tip — The minus sign applies to the entire second integral. Expanding -2(xe^x-e^x) gives -2xe^x+2e^x.

4

Definite integrals by parts

Use ∫abuv′ dx=[uv]ab−∫abvu′ dx\int_a^b u v'\,dx=[uv]_a^b-\int_a^b vu'\,dx. Both boundary terms and the remaining integral use the original limits. A definite integral gives a value, so there is no constant of integration in the final answer.

Example 1Evaluate ∫0π/2xcos⁡x dx\int_0^{\pi/2}x\cos x\,dx.
1Choose u=xu=x, dvdx=cos⁡x\tfrac{dv}{dx}=\cos x, giving dudx=1\tfrac{du}{dx}=1 and v=sin⁡xv=\sin x.
2[xsin⁡x]0π/2−∫0π/2sin⁡x dx[x\sin x]_0^{\pi/2}-\int_0^{\pi/2}\sin x\,dx.
3The boundary term is π2×1−0=π2\tfrac{\pi}{2}\times1-0=\tfrac{\pi}{2}; the remaining integral is [−cos⁡x]0π/2=1[-\cos x]_0^{\pi/2}=1.
Answer
π2−1\dfrac{\pi}{2}-1

Formula recap

∫u v′ dx=uv−∫v u′ dx\int u\,v'\,dx = uv - \int v\,u'\,dx
By parts.
∫ln⁡x dx=xln⁡x−x+c\int \ln x\,dx = x\ln x - x + c
Special case.

Common mistakes to avoid

Choosing u as the part that gets more complicated.
Pick u so its derivative simplifies (use LATE order).
Dropping the minus sign before the second integral.
It is uv − ∫v u′ dx.

Key takeaways

  • ∫u v′ dx = uv − ∫v u′ dx.
  • Choose u by LATE (logs first, exponentials last).
  • ∫ln x dx = x ln x − x + c.

Test yourself

Answer a few questions, then review anything you missed.

Prefer paper? Open the written self-check

Answer each question before opening its answer. This self-check does not record a quiz score.

  1. ∫xsin⁡x dx\int x\sin x\,dx equals:

    • −xcos⁡x+sin⁡x+c-x\cos x+\sin x+c
    • xcos⁡x−sin⁡x+cx\cos x-\sin x+c
    • −xcos⁡x−sin⁡x+c-x\cos x-\sin x+c
    • xsin⁡x+cos⁡x+cx\sin x+\cos x+c
    Show answer and explanation

    Answer: −xcos⁡x+sin⁡x+c-x\cos x+\sin x+c

    Take u=x and v=-cos x. Then uv minus the integral of v gives -x cos x + sin x + c. Differentiating confirms x sin x.

  2. Integration by parts reverses the:

    • product rule
    • chain rule
    • quotient rule
    • power rule
    Show answer and explanation

    Answer: product rule

    Its origin.

  3. The formula is ∫u v′ dx =

    • uv−∫vu′ dxuv - \int v u'\,dx
    • uv+∫vu′ dxuv + \int v u'\,dx
    • u′v−uv′u'v - uv'
    • ∫uv dx\int uv\,dx
    Show answer and explanation

    Answer: uv−∫vu′ dxuv - \int v u'\,dx

    Standard.

  4. For ∫xexdx\int x e^x dx, choose u =

    • xx
    • exe^x
    • xexx e^x
    • 11
    Show answer and explanation

    Answer: xx

    Algebra before exponential.

  5. ∫xex dx\int x e^x\,dx equals:

    • ex(x−1)+ce^x(x-1)+c
    • xex+cx e^x + c
    • ex(x+1)+ce^x(x+1)+c
    • x22ex+c\tfrac{x^2}{2}e^x+c
    Show answer and explanation

    Answer: ex(x−1)+ce^x(x-1)+c

    xeˣ − eˣ.

  6. ∫ln⁡x dx\int \ln x\,dx equals:

    • xln⁡x−x+cx\ln x - x + c
    • 1x+c\frac{1}{x}+c
    • xln⁡x+cx\ln x + c
    • ln⁡x+c\ln x + c
    Show answer and explanation

    Answer: xln⁡x−x+cx\ln x - x + c

    Parts with dv/dx = 1.

Next step

Check your understanding

Work through questions on Integration by Parts. Use the feedback to decide what needs another look.

1 past-paper question · 4 marks

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