P4 6.2 - Integration by parts | Edexcel A-Level International Mathematics

Reverse the product rule to integrate products whose factors cannot be handled by substitution. Learn how to choose a useful factor to differentiate, integrate logarithms, and continue when the method must be applied more than once.

Reversing the product rule

Can you integrate $xe^x$ by integrating its two factors separately? The guess $(x^2/2)e^x$ fails: its derivative contains an unwanted extra term. Instead, start from a product whose derivative includes $xe^x$ .

The product rule gives $(xe^x)'=e^x+xe^x$ . Rearrange and integrate to obtain $\int xe^x\,dx=xe^x-\int e^x\,dx$ . The difficult product has become a product minus a simpler integral.

Integration by parts

For differentiable functions $u$ and $v$ ,

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

Choose $u$ , differentiate it, choose $dv/dx$ , and integrate that to obtain $v$ .

In the formula booklet.

The formula follows by integrating $(uv)'=u'v+uv'$ and rearranging. Its minus sign is essential.

Worked example 1: remove a factor of x

Find $\int xe^x\,dx$ .

Plan. Differentiate $u=x$ so that it becomes $1$ , and integrate $dv/dx=e^x$ .

u=x,\quad \frac{du}{dx}=1,\quad \frac{dv}{dx}=e^x,\quad v=e^x.

\int xe^x\,dx=xe^x-\int e^x\,dx=xe^x-e^x+C.

Answer. $e^x(x-1)+C$ .

Check. Differentiating the factorised answer gives $e^x(x-1)+e^x=xe^x$ .

Change only the exponential's inner coefficient from $1$ to $2$ . Every integration of the exponential now needs a factor of $1/2$ .

Worked example 2: keep the integration factor

Find $\int xe^{2x}\,dx$ .

Plan. Keep $u=x$ , but use $v=\frac12e^{2x}$ .

\begin{aligned} \int xe^{2x}\,dx &=\frac{x}{2}e^{2x}-\frac12\int e^{2x}\,dx\\ &=\frac{x}{2}e^{2x}-\frac14e^{2x}+C \end{aligned}

Answer. $\frac14e^{2x}(2x-1)+C$ .

Check. Its derivative is $\frac12e^{2x}(2x-1)+\frac12e^{2x}=xe^{2x}$ .

Choose $v$ without an arbitrary constant; one final $C$ covers the family of antiderivatives.

Watch out

For $dv/dx=e^{2x}$ , $v=\frac12e^{2x}$ , not $2e^{2x}$ . Check this small integration before inserting it into the parts formula.

Only the exponential coefficient changes again.

Choosing the differentiated factor

A useful choice makes the remaining integral easier than the original one. A polynomial usually simplifies when differentiated, while an exponential, sine or cosine stays easy to integrate.

For $x\cos x$ , choosing $u=x$ leaves only sine to integrate. Choosing $u=\cos x$ instead gives a remaining integral involving $x^2\sin x$ : valid algebra, but harder work. A memorised ordering can suggest a choice; the decisive test is whether the new integral becomes simpler.

Worked example 3: a trigonometric product

Find $\int x\cos x\,dx$ .

Plan. Set $u=x$ and $dv/dx=\cos x$ , so $v=\sin x$ .

\int x\cos x\,dx=x\sin x-\int\sin x\,dx =x\sin x+\cos x+C.

The final plus sign comes from subtracting the antiderivative $-\cos x$ .

Answer. $x\sin x+\cos x+C$ .

Check. Differentiation gives $\sin x+x\cos x-\sin x=x\cos x$ .

Change only cosine to sine. The best differentiated factor remains $x$ , but the integrated factor has a minus sign.

Worked example 4: two minus signs

Find $\int x\sin x\,dx$ .

Plan. Choose $u=x$ , $du/dx=1$ and $v=-\cos x$ .

\begin{aligned} \int x\sin x\,dx &=-x\cos x-\int(-\cos x)\,dx\\ &=-x\cos x+\sin x+C \end{aligned}

Answer. $-x\cos x+\sin x+C$ .

Check. The derivative is $-\cos x+x\sin x+\cos x=x\sin x$ .

For definite integrals, evaluate the product at both limits as well as the remaining integral: $\int_a^buv'\,dx=[uv]_a^b-\int_a^bvu'\,dx$ . Equivalently, first find the complete antiderivative, then substitute the limits.

Problem-solving

Before calculating, mentally compare the two choices for $u$ . Choose the one that reduces a polynomial's degree or changes a logarithm into a simpler reciprocal.

Keep this choice and change only the angle multiplier.

Integrating logarithms

The single function $\ln x$ can be treated as the product $(\ln x)\times1$ . Differentiate the logarithm, because $1/x$ cancels the $x$ obtained by integrating $1$ .

The domain $x>0$ is needed for $\ln x$ . If the integrand were $\ln|x|$ , the same reasoning would apply on either interval avoiding zero.

Worked example 5: supply a factor of one

Find $\int\ln x\,dx$ , for $x>0$ .

Plan. Choose $u=\ln x$ , $du/dx=1/x$ , $dv/dx=1$ , $v=x$ .

\int\ln x\,dx=x\ln x-\int x\frac1x\,dx =x\ln x-x+C.

The remaining integrand is $1$ , not $0$ .

Answer. $x\ln x-x+C$ .

Check. Differentiating gives $\ln x+1-1=\ln x$ .

Change only the factor multiplying $\ln x$ , from $1$ to $x$ . Keep differentiating the logarithm.

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P4 6.2 - Integration by parts

Reversing the product rule

Choosing the differentiated factor

Integrating logarithms

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