A-level Maths revision · Pure, year 2

Integration by parts

The idea

∫udvdx dx=uv−∫vdudx dx\int u\frac{dv}{dx}\,dx = uv - \int v\frac{du}{dx}\,dx. Choose uu to be the part that gets simpler when differentiated, usually the power of xx; with ln⁡x\ln x, always choose u=ln⁡xu = \ln x.

  1. Pick uu and dvdx\frac{dv}{dx}; find dudx\frac{du}{dx} and vv.
  2. Write uv−∫vdudx dxuv - \int v\frac{du}{dx}\,dx.
  3. Integrate what is left; use parts again if needed.

Watch out Keep the minus sign in front of the second integral.

See it worked

Worked example

Find ∫xex dx\int xe^x\,dx.

  1. u=xu = x, dvdx=ex\frac{dv}{dx} = e^x; so dudx=1\frac{du}{dx} = 1, v=exv = e^x.
  2. xex−∫ex dxxe^x - \int e^x\,dx.
  3. =xex−ex+c= xe^x - e^x + c.

Answer xex−ex+cxe^x - e^x + c

Your turn

Question 1Non-calculator3 marks

Find ∫xcos⁡x dx\int x\cos x\,dx.

Question 2Non-calculator3 marks

Find ∫ln⁡x dx\int \ln x\,dx.

Question 3Non-calculator3 marks

Find the exact value of ∫01xe2x dx\int_0^1 xe^{2x}\,dx.

In past papers

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