8.6 Integrating with partial fractions — Practise

Work through these on your own. Factorise the denominator, split the fraction, then integrate each part. Check any improper fraction for division first.

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

Find $\displaystyle\int \frac{1}{(x - 1)(x + 2)}\,dx$ .
Find $\displaystyle\int \frac{5x + 1}{(x - 1)(x + 2)}\,dx$ .
Find $\displaystyle\int \frac{2x + 3}{(x + 1)(x + 2)}\,dx$ .
Find $\displaystyle\int \frac{x + 4}{(x - 2)(x + 1)}\,dx$ .
Find $\displaystyle\int \frac{3x - 2}{(x + 1)(x - 3)}\,dx$ .
Find $\displaystyle\int \frac{1}{x(x - 1)^2}\,dx$ .
Evaluate $\displaystyle\int_2^3 \frac{1}{(x - 1)(x + 2)}\,dx$ , giving an exact answer.
Find $\displaystyle\int \frac{x^2 + 1}{x(x - 1)}\,dx$ .
Find $\displaystyle\int \frac{1}{(x + 1)(x + 3)}\,dx$ .
Find $\displaystyle\int \frac{1}{x(x + 1)^2}\,dx$ .
Find $\displaystyle\int \frac{x^2}{x^2 - 1}\,dx$ .
Evaluate $\displaystyle\int_0^1 \frac{1}{(x + 1)(x + 2)}\,dx$ , giving an exact answer.

Answers and concise solutions

1. The integral

Answer: Split into $\dfrac{1}{3(x - 1)} - \dfrac{1}{3(x + 2)}$ . Integrating gives $\dfrac{1}{3}\ln|x - 1| - \dfrac{1}{3}\ln|x + 2| + C = \dfrac{1}{3}\ln\left|\dfrac{x - 1}{x + 2}\right| + C$ .

2. The integral

Answer: The split is $\dfrac{2}{x - 1} + \dfrac{3}{x + 2}$ , so the integral is $2\ln|x - 1| + 3\ln|x + 2| + C$ .

3. The integral

Answer: The split is $\dfrac{1}{x + 1} + \dfrac{1}{x + 2}$ , so the integral is $\ln|x + 1| + \ln|x + 2| + C = \ln|(x + 1)(x + 2)| + C$ .

4. The integral

Answer: The split is $\dfrac{2}{x - 2} - \dfrac{1}{x + 1}$ , so the integral is $2\ln|x - 2| - \ln|x + 1| + C$ .

5. The integral

Answer: The split is $\dfrac{5}{4(x + 1)} + \dfrac{7}{4(x - 3)}$ , so the integral is $\dfrac{5}{4}\ln|x + 1| + \dfrac{7}{4}\ln|x - 3| + C$ .

6. The integral

Answer: The split is $\dfrac{1}{x} - \dfrac{1}{x - 1} - \dfrac{1}{(x - 1)^2}$ . Integrating gives $\ln|x| - \ln|x - 1| - \dfrac{1}{x - 1} + C$ .

7. The value of the integral

Answer: The antiderivative is $\dfrac{1}{3}\ln\left|\dfrac{x - 1}{x + 2}\right|$ . Then $\dfrac{1}{3}\left(\ln\dfrac{2}{5} - \ln\dfrac{1}{4}\right) = \dfrac{1}{3}\ln\dfrac{8}{5}$ .

8. The integral

Answer: Divide first: $\dfrac{x^2 + 1}{x(x - 1)} = 1 - \dfrac{1}{x} + \dfrac{2}{x - 1}$ . Integrating gives $x - \ln|x| + 2\ln|x - 1| + C$ .

9. The integral

Answer: The split is $\dfrac{1}{2(x + 1)} - \dfrac{1}{2(x + 3)}$ , so the integral is $\dfrac{1}{2}\ln|x + 1| - \dfrac{1}{2}\ln|x + 3| + C$ .

10. The integral

Answer: The split is $\dfrac{1}{x} - \dfrac{1}{x + 1} - \dfrac{1}{(x + 1)^2}$ . Integrating gives $\ln|x| - \ln|x + 1| + \dfrac{1}{x + 1} + C$ .

11. The integral

Answer: Divide first: $\dfrac{x^2}{x^2 - 1} = 1 + \dfrac{1}{2(x - 1)} - \dfrac{1}{2(x + 1)}$ . Integrating gives $x + \dfrac{1}{2}\ln\left|\dfrac{x - 1}{x + 1}\right| + C$ .

12. The value of the integral

Answer: The integral is $\Big[\ln(x + 1) - \ln(x + 2)\Big]_0^1 = (\ln 2 - \ln 3) + \ln 2 = \ln\dfrac{4}{3}$ .