8.6 Integrating with partial fractions — Practise with help

Work each question closed book first. Split the fraction completely before you integrate, and write the constant for each partial fraction in the numerator.

Warm-up: remember

Write $\dfrac{3}{x - 1} + \dfrac{2}{x + 1}$ as a single fraction.
Simplify $\ln 8 - \ln 2$ .
Find $\displaystyle\int \frac{1}{x}dx$ .

Answers

$\dfrac{3(x + 1) + 2(x - 1)}{(x - 1)(x + 1)} = \dfrac{5x + 1}{(x - 1)(x + 1)}$ .
$\ln 8 - \ln 2 = \ln 4 = 2\ln 2$ .
$\ln x + C$ for $x > 0$ .

If question 1 was hard, revise partial fractions in Section 2.8.

Practise with help

1. Find $\displaystyle\int \frac{1}{(x + 1)(x + 3)}dx$ .

Hint

Write the fraction as $\dfrac{A}{x + 1} + \dfrac{B}{x + 3}$ , then substitute $x = -1$ and $x = -3$ .

Next step

$1 = A(x + 3) + B(x + 1)$ . At $x = -1$ , $A = \dfrac{1}{2}$ . At $x = -3$ , $B = -\dfrac{1}{2}$ .

Full working

\int \frac{1}{(x + 1)(x + 3)}dx = \frac{1}{2}\ln|x + 1| - \frac{1}{2}\ln|x + 3| + C.

Both terms are logarithms, because each partial fraction has a power of one in the denominator.

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

Hint

The numerator is not a constant, so the two constants $A$ and $B$ need not be equal. Use $2x + 3 = A(x + 2) + B(x + 1)$ .

Next step

At $x = -1$ : $1 = A$ . At $x = -2$ : $-1 = -B$ , so $B = 1$ .

Full working

\frac{2x + 3}{(x + 1)(x + 2)} = \frac{1}{x + 1} + \frac{1}{x + 2}.

\int \frac{2x + 3}{(x + 1)(x + 2)}dx = \ln|x + 1| + \ln|x + 2| + C = \ln|(x + 1)(x + 2)| + C.

The two logarithms combine with the law $\ln a + \ln b = \ln ab$ .

3. Find $\displaystyle\int \frac{x + 4}{(x - 2)(x + 1)}dx$ .

Hint

Use $x + 4 = A(x + 1) + B(x - 2)$ . Substitute the values that make each factor zero.

Next step

At $x = 2$ : $6 = 3A$ , so $A = 2$ . At $x = -1$ : $3 = -3B$ , so $B = -1$ .

Full working

\int \frac{x + 4}{(x - 2)(x + 1)}dx = 2\ln|x - 2| - \ln|x + 1| + C.

The sign of each coefficient is the sign in the final answer.

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

Hint

The repeated factor needs two fractions, for $(x + 1)$ and for $(x + 1)^2$ : $\dfrac{A}{x} + \dfrac{B}{x + 1} + \dfrac{C}{(x + 1)^2}$ .

Next step

$1 = A(x + 1)^2 + Bx(x + 1) + Cx$ . At $x = 0$ , $A = 1$ . At $x = -1$ , $C = -1$ . Comparing the $x^2$ coefficients gives $0 = A + B$ , so $B = -1$ .

Full working

\int \frac{1}{x(x + 1)^2}dx = \ln|x| - \ln|x + 1| + \frac{1}{x + 1} + C.

The fraction with the squared denominator integrates with the power rule. Its negative coefficient $C = -1$ and the negative power combine to give a positive term.

5. Find $\displaystyle\int \frac{x^2}{x^2 - 1}dx$ .

Hint

The top and bottom have the same degree, so divide first. Write $\dfrac{x^2}{x^2 - 1} = 1 + \dfrac{1}{x^2 - 1}$ .

Next step

Split $\dfrac{1}{(x - 1)(x + 1)} = \dfrac{1/2}{x - 1} - \dfrac{1/2}{x + 1}$ .

Full working

\int \frac{x^2}{x^2 - 1}dx = \int 1dx + \frac{1}{2}\int \frac{1}{x - 1}dx - \frac{1}{2}\int \frac{1}{x + 1}dx = x + \frac{1}{2}\ln\left|\frac{x - 1}{x + 1}\right| + C.

Dividing first removed the improper part and left a fraction with linear factors.

6. Evaluate $\displaystyle\int_0^1 \frac{1}{(x + 1)(x + 2)}dx$ , giving an exact answer.

Hint

Split the fraction first. The two linear factors are $(x + 1)$ and $(x + 2)$ .

Next step

$\dfrac{1}{(x + 1)(x + 2)} = \dfrac{1}{x + 1} - \dfrac{1}{x + 2}$ . Integrate to $\ln|x + 1| - \ln|x + 2|$ .

Full working

\int_0^1 \frac{1}{(x + 1)(x + 2)}dx = \Big[\ln(x + 1) - \ln(x + 2)\Big]_0^1.

At the upper limit: $\ln 2 - \ln 3$ . At the lower limit: $\ln 1 - \ln 2 = -\ln 2$ . Subtracting,

(\ln 2 - \ln 3) - (-\ln 2) = 2\ln 2 - \ln 3 = \ln\frac{4}{3}.

Both logarithms are positive on the interval, so no absolute values are needed.