8.5 Integration by substitution and by parts — Practise

Work through these on your own. For each integral, decide first whether a substitution or parts is needed, then name the parts you will use. Check each indefinite integral by differentiating your answer.

The questions and the answers below come from a set that is checked by computer every time the book is built.

Find $\displaystyle\int 2x(x^2 + 1)^4\,dx$ .
Find $\displaystyle\int 3x^2e^{x^3}\,dx$ .
Find $\displaystyle\int \frac{2x}{x^2 + 1}\,dx$ .
Find $\displaystyle\int x(x^2 + 3)^5\,dx$ .
Find $\displaystyle\int xe^{2x}\,dx$ .
Find $\displaystyle\int x\sin x\,dx$ .
Find $\displaystyle\int x\ln x\,dx$ .
Find $\displaystyle\int \ln x\,dx$ .
Find $\displaystyle\int x^2e^x\,dx$ .
Evaluate $\displaystyle\int_0^1 xe^x\,dx$ .
Evaluate $\displaystyle\int_0^1 2x(x^2 + 1)^3\,dx$ .
Find $\displaystyle\int \cos^3 x\sin x\,dx$ .

Answers and concise solutions

1. The integral

Answer: Let $u = x^2 + 1$ , so $du = 2x\,dx$ . Then $\displaystyle\int u^4\,du = \dfrac{u^5}{5} + C = \dfrac{(x^2 + 1)^5}{5} + C$ .

2. The integral

Answer: Let $u = x^3$ , so $du = 3x^2\,dx$ . Then $\displaystyle\int e^u\,du = e^u + C = e^{x^3} + C$ .

3. The integral

Answer: Let $u = x^2 + 1$ , so $du = 2x\,dx$ . Then $\displaystyle\int \dfrac{1}{u}\,du = \ln u + C = \ln(x^2 + 1) + C$ .

4. The integral

Answer: Let $u = x^2 + 3$ , so $x\,dx = \dfrac{1}{2}du$ . Then $\dfrac{1}{2}\displaystyle\int u^5\,du = \dfrac{u^6}{12} + C$ .

5. The integral

Answer: By parts with $u = x$ and $v = \dfrac{1}{2}e^{2x}$ : $uv - \displaystyle\int v\,du = \dfrac{1}{2}xe^{2x} - \dfrac{1}{4}e^{2x} + C = \dfrac{e^{2x}(2x - 1)}{4} + C$ .

6. The integral

Answer: By parts with $u = x$ and $v = -\cos x$ : $-x\cos x + \displaystyle\int \cos x\,dx = -x\cos x + \sin x + C$ .

7. The integral

Answer: By parts with $u = \ln x$ and $v = \dfrac{x^2}{2}$ : $\dfrac{x^2}{2}\ln x - \displaystyle\int \dfrac{x}{2}\,dx = \dfrac{x^2}{2}\ln x - \dfrac{x^2}{4} + C$ .

8. The integral

Answer: Write $\ln x$ as $\ln x \times 1$ , and use parts with $u = \ln x$ and $v = x$ : $x\ln x - \displaystyle\int 1\,dx = x\ln x - x + C$ .

9. The integral

Answer: Apply parts twice. First $\displaystyle\int x^2e^x\,dx = x^2e^x - 2\displaystyle\int xe^x\,dx$ , and then $\displaystyle\int xe^x\,dx = xe^x - e^x$ . So the integral is $e^x(x^2 - 2x + 2) + C$ .

10. The value of the integral

Answer: By parts, an antiderivative is $xe^x - e^x$ . Then $\Big[xe^x - e^x\Big]_0^1 = (e - e) - (0 - 1) = 1$ .

11. The value of the integral

Answer: Let $u = x^2 + 1$ , so the limits become $u = 1$ and $u = 2$ . Then $\displaystyle\int_1^2 u^3\,du = \left[\dfrac{u^4}{4}\right]_1^2 = \dfrac{15}{4}$ .

12. The integral

Answer: Let $u = \cos x$ , so $du = -\sin x\,dx$ . Then $-\displaystyle\int u^3\,du = -\dfrac{u^4}{4} + C = -\dfrac{\cos^4 x}{4} + C$ .