8.5 Integration by substitution and by parts — Practise with help

Work each question closed book first. For a substitution, write $u$ and $du$ before you change the integral. For parts, write $u$ and $\dfrac{dv}{dx}$ .

Warm-up: remember

Differentiate $y = (x^2 + 1)^4$ .
Differentiate $y = xe^{2x}$ .
Find $\displaystyle\int e^{3x}dx$ .

Answers

$\dfrac{dy}{dx} = 4(x^2 + 1)^3 \times 2x = 8x(x^2 + 1)^3$ .
$\dfrac{dy}{dx} = e^{2x} + 2xe^{2x} = e^{2x}(1 + 2x)$ .
$\dfrac{1}{3}e^{3x} + C$ .

If question 1 was hard, revise the chain rule in Section 7.7.

Practise with help

1. Find $\displaystyle\int 2x(x^2 + 1)^4dx$ .

Hint

Let $u = x^2 + 1$ . What is $du$ ? Look for $2xdx$ in the integral.

Next step

With $u = x^2 + 1$ , $du = 2xdx$ . The integral becomes $\displaystyle\int u^4du$ .

Full working

\int u^4du = \frac{u^5}{5} + C.

Substitute back:

\int 2x(x^2 + 1)^4dx = \frac{(x^2 + 1)^5}{5} + C.

Check by differentiating the answer with the chain rule.

2. Find $\displaystyle\int 3x^2e^{x^3}dx$ .

Hint

The derivative of $x^3$ is $3x^2$ , which appears in the integral. Let $u = x^3$ .

Next step

$du = 3x^2dx$ , so the integral becomes $\displaystyle\int e^udu$ .

Full working

\int e^udu = e^u + C = e^{x^3} + C.

The substitution removes the $3x^2$ and leaves the standard exponential integral.

3. Find $\displaystyle\int \frac{2x}{x^2 + 1}dx$ .

Hint

Let $u = x^2 + 1$ . The top is the derivative of the bottom, which points to a logarithm.

Next step

$du = 2xdx$ , so the integral becomes $\displaystyle\int \frac{1}{u}du$ .

Full working

\int \frac{1}{u}du = \ln u + C = \ln(x^2 + 1) + C.

The quadratic is always positive, so the logarithm is defined for every $x$ .

4. Find $\displaystyle\int xe^{2x}dx$ using integration by parts.

Hint

Take $u = x$ , so that differentiating removes the $x$ . Take $\dfrac{dv}{dx} = e^{2x}$ .

Next step

$\dfrac{du}{dx} = 1$ and $v = \dfrac{1}{2}e^{2x}$ . Write $uv - \displaystyle\int v\frac{du}{dx}dx$ .

Full working

\int xe^{2x}dx = x \times \frac{1}{2}e^{2x} - \int \frac{1}{2}e^{2x}dx = \frac{1}{2}xe^{2x} - \frac{1}{4}e^{2x} + C = \frac{e^{2x}(2x - 1)}{4} + C.

The second integral is simpler than the first because the $x$ is gone.

5. Find $\displaystyle\int x\cos xdx$ using integration by parts.

Hint

Take $u = x$ and $\dfrac{dv}{dx} = \cos x$ . The integral of cosine is sine.

Next step

$\dfrac{du}{dx} = 1$ and $v = \sin x$ . Write $uv - \displaystyle\int vdu$ .

Full working

\int x\cos xdx = x\sin x - \int \sin xdx = x\sin x + \cos x + C.

The minus sign in the formula turns $-\sin x$ into $+\cos x$ after integration.

6. Find $\displaystyle\int x^2e^xdx$ .

Hint

Apply integration by parts with $u = x^2$ . The new integral still contains $x$ , so parts is needed a second time.

Next step

With $u = x^2$ and $v = e^x$ , $\displaystyle\int x^2e^xdx = x^2e^x - 2\int xe^xdx$ . Now find $\displaystyle\int xe^xdx$ .

Full working

The second integral is

\int xe^xdx = xe^x - \int e^xdx = xe^x - e^x.

Substitute this back:

\int x^2e^xdx = x^2e^x - 2\left(xe^x - e^x\right) + C = e^x\left(x^2 - 2x + 2\right) + C.

Apply parts twice whenever the power of $x$ is greater than one.