H5 - Integration by parts | AQA A-Level Mathematics

Integrate products by reversing the product rule. You will choose which factor to differentiate, integrate logarithms, apply the method twice when needed and keep both boundary terms in definite integrals.

Reversing the product rule

How can you integrate $xe^{2x}$ when neither factor is the derivative of the other's inside? Differentiating $\frac12xe^{2x}$ gives $xe^{2x}+\frac12e^{2x}$ . The extra term is easy to integrate and subtract. This is the idea behind integration by parts.

You already know

The product rule gives $\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$ . Integrating $e^{ax}$ , for constant $a\ne0$ , gives $e^{ax}/a+C$ .

Integrate the product rule and rearrange: $uv=\int u\frac{dv}{dx}\,dx+\int v\frac{du}{dx}\,dx$ . Moving the second integral to the other side explains the minus sign. Arbitrary constants can be collected into one final constant.

Integration by parts

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

Choose $u$ to differentiate and $dv/dx$ to integrate, on an interval where the functions and integrals are defined.
In the formula booklet.

A useful choice simplifies the remaining integral. Here differentiating the polynomial leaves a constant, whereas integrating the exponential keeps the same kind of function. The exponential antiderivative recalled above is not in the formula booklet: learn it.

Worked example 1: a linear factor disappears

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

Plan. Differentiate $x$ , integrate $e^{2x}$ , then integrate the simpler remainder.

\begin{aligned} u&=x,\quad \frac{du}{dx}=1 && \text{differentiate the polynomial} \\ \frac{dv}{dx}&=e^{2x},\quad v=\frac12e^{2x} && \text{integrate the exponential} \\ \int xe^{2x}\,dx&=\frac12xe^{2x}-\frac12\int e^{2x}\,dx && \text{parts} \\ &=\frac12xe^{2x}-\frac14e^{2x}+C && \text{integrate again} \end{aligned}

Why this step? Differentiating $x$ removes the awkward factor, so the remaining integral is a single exponential.

Answer. $\frac12xe^{2x}-\frac14e^{2x}+C$ .

Check. Differentiation gives $\frac12e^{2x}+xe^{2x}-\frac12e^{2x}=xe^{2x}$ .

Change only the coefficient of $x$ in the exponential, from $2$ to $3$ . Each integration of that exponential now contributes a factor $1/3$ .

Worked example 2: keep the chain factors

Question. Find $\int xe^{3x}\,dx$ .

Plan. Use the same factor choices, keeping the new factor $1/3$ at both integrations.

\begin{aligned} u&=x,\quad \frac{du}{dx}=1 && \text{differentiate} \\ \frac{dv}{dx}&=e^{3x},\quad v=\frac13e^{3x} && \text{reverse chain rule} \\ \int xe^{3x}\,dx&=\frac13xe^{3x}-\frac13\int e^{3x}\,dx && \text{parts} \\ &=\frac13xe^{3x}-\frac19e^{3x}+C && \text{integrate remainder} \end{aligned}

Why this step? The coefficient outside the remaining integral multiplies the factor from integrating inside it; $(1/3)(1/3)=1/9$ .

Answer. $\frac13xe^{3x}-\frac19e^{3x}+C$ .

Check. The derivative is $\frac13e^{3x}+xe^{3x}-\frac13e^{3x}=xe^{3x}$ .

You can take the constant in $v$ as zero. Adding a constant to $v$ cancels between the product term and the remaining integral, up to the final arbitrary constant.

Watch out

Wrong: $\int xe^{3x}\,dx=\frac13xe^{3x}+\frac19e^{3x}+C$ . The product rule's extra term must be subtracted. The correct second term is $-e^{3x}/9$ .

Change only the exponential coefficient again in the next question. Write enough working to make the factor choices visible.

Choosing factors and integrating logarithms

A logarithm is often the factor to differentiate because its derivative is a reciprocal. For $\ln x$ , write an otherwise hidden product: $(\ln x)\times1$ . Integrating the $1$ produces $x$ , which cancels the reciprocal $1/x$ in the remainder.

The domain of $\ln x$ is $x>0$ . Its derivative $1/x$ and the power rule are not in the formula booklet: learn them. The integration-by-parts rule is supplied; the antiderivatives you derive here are not separately printed.

Worked example 3: reveal a factor of one

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

Plan. Differentiate the logarithm and integrate its factor $1$ .

\begin{aligned} u&=\ln x,\quad \frac{du}{dx}=\frac1x && \text{differentiate the logarithm} \\ \frac{dv}{dx}&=1,\quad v=x && \text{integrate the other factor} \\ \int\ln x\,dx&=x\ln x-\int x\frac1x\,dx && \text{parts} \\ &=x\ln x-x+C && \text{integrate }1 \end{aligned}

Why this step? The reciprocal derivative cancels the new factor $x$ , leaving no logarithm in the remaining integral.

Answer. $x\ln x-x+C$ , for $x>0$ .

Check. $\frac{d}{dx}(x\ln x-x)=\ln x+1-1=\ln x$ .

Change only the polynomial factor, from $1$ to $x$ . Keeping the logarithm as $u$ still makes the remaining integral a polynomial.

Worked example 4: differentiate the logarithm again

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

Plan. Differentiate $\ln x$ and integrate $x$ .

\begin{aligned} u&=\ln x,\quad \frac{du}{dx}=\frac1x && \text{same choice of }u \\ \frac{dv}{dx}&=x,\quad v=\frac12x^2 && \text{power rule} \\ \int x\ln x\,dx&=\frac12x^2\ln x-\frac12\int x\,dx && \text{cancel one }x \\ &=\frac12x^2\ln x-\frac14x^2+C && \text{integrate remainder} \end{aligned}

Why this step? Although $x$ also simplifies when differentiated, differentiating the logarithm removes it altogether from the remainder.

Answer. $\frac12x^2\ln x-\frac14x^2+C$ , for $x>0$ .

Check. Its derivative is $x\ln x+x/2-x/2=x\ln x$ .

Choosing $u=1$ in Worked example 3 would require integrating $\ln x$ to find $v$ , which is exactly the task you started with. A choice is useful when you can integrate the other factor and the remainder becomes simpler. For trig products such as $x\cos3x$ , differentiating $x$ and integrating the cosine is useful instead.

Watch out

Wrong: $\int\ln x\,dx=\ln x\int1\,dx=x\ln x+C$ . The factor $\ln x$ is not constant. Parts subtracts the missing correction $\int1\,dx=x$ , giving $x\ln x-x+C$ .

Increase just the power of the polynomial factor in the next question. The reciprocal derivative still removes one power of $x$ .

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H5 - Integration by parts

Reversing the product rule

Choosing factors and integrating logarithms

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