WAEC 2016 · Paper 2 · Q9

(a)
Express $\dfrac{x + 6}{(x + 1)^3}$ in partial fractions.
(b)
Use the answer in (a) to evaluate $\displaystyle\int_1^2 \frac{x + 6}{(x + 1)^3}\,dx$ .

Worked solution (try it first)

(a)

Write $\dfrac{A}{x + 1} + \dfrac{B}{(x + 1)^2} + \dfrac{C}{(x + 1)^3}$ and multiply through: $x + 6 = A(x + 1)^2 + B(x + 1) + C$ .
Put $x = -1$ : $5 = C$ .
The $x^2$ terms: $0 = A$ .
The $x$ terms: $1 = 2A + B$ , so $B = 1$ .
So the answer is $\dfrac{1}{(x + 1)^2} + \dfrac{5}{(x + 1)^3}$ .

(b)

Integrate each piece: $\displaystyle\int (x + 1)^{-2}\,dx = -\frac{1}{x + 1}$ and $\displaystyle\int 5(x + 1)^{-3}\,dx = -\frac{5}{2(x + 1)^2}$ .
At $x = 2$ : $-\frac13 - \frac{5}{18} = -\frac{11}{18}$ .
At $x = 1$ : $-\frac12 - \frac58 = -\frac98$ .
Subtract: $-\frac{11}{18} + \frac98 = \frac{-44 + 81}{72}$
$= \frac{37}{72}$
$\approx 0.514$ .

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