CAIE A-Level · Mathematics 9709 · Integration

Integration by Parts (Pure Mathematics 3 – 9709)

9 min readSyllabus 3.5PreviewBy Uzair Khan

Syllabus objective

Recognise when an integrand can usefully be regarded as a product, and use integration by parts, e.g. integration of x sin 2x, x² e^(−x), ln x, x tan⁻¹ x.

Introduction

Integration by parts is one of the most important techniques in Pure Mathematics 3. It allows you to integrate products of functions that cannot be handled by standard forms or substitution alone. The syllabus specifically requires you to recognise when an integrand should be treated as a product, and to apply the technique confidently to expressions such as xsin2xx \sin 2x, x2exx^2 e^{-x}, lnx\ln x, and xtan1xx \tan^{-1} x. Questions involving this method appear regularly in Paper 3, often combined with definite integration or asking for an exact answer.


Core Concept

The method is derived from the product rule for differentiation. If uu and vv are functions of xx, the product rule states:

ddx(uv)=udvdx+vdudx\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}

Rearranging and integrating both sides with respect to xx gives the integration by parts formula:

udvdxdx=uvvdudxdx\int u \frac{dv}{dx}\, dx = uv - \int v \frac{du}{dx}\, dx

Choosing uu and dvdx\dfrac{dv}{dx} is the critical skill. Use the LIATE priority order as a guide — whichever function type appears earliest in this list should be chosen as uu:

PriorityTypeExamples
1stLogarithmslnx\ln x, logax\log_a x
2ndInverse trigtan1x\tan^{-1} x, sin1x\sin^{-1} x
3rdAlgebraic (polynomial)xx, x2x^2, x3x^3
4thTrigonometricsinx\sin x, cosx\cos x
5thExponentialexe^x, exe^{-x}

The key principle: choose uu so that dudx\dfrac{du}{dx} is simpler than uu, and choose dvdx\dfrac{dv}{dx} so that you can integrate it to find vv.

Special cases: For lnx\ln x and tan1x\tan^{-1} x alone (apparently not a product), write the integrand as 1lnx1 \cdot \ln x or 1tan1x1 \cdot \tan^{-1} x and set u=lnxu = \ln x (or tan1x\tan^{-1} x) and dvdx=1\dfrac{dv}{dx} = 1.

Repeated application: For integrands like x2exx^2 e^{-x}, you must apply integration by parts twice, each time reducing the power of the polynomial factor.


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