WAEC 2017 · Paper 2 · Q2

Worked solution (try it first)

Split into partial fractions: $\dfrac{x - 1}{(x + 1)^2} = \dfrac{A}{x + 1} + \dfrac{B}{(x + 1)^2}$ , so $x - 1 = A(x + 1) + B$ .
Put $x = -1$ : $B = -2$ .
Compare the $x$ terms: $A = 1$ .
So the integral is $\displaystyle\int_1^3 \left(\frac{1}{x + 1} - \frac{2}{(x + 1)^2}\right)dx$ .
Integrate each piece: $\left[\ln(x + 1) + \dfrac{2}{x + 1}\right]_1^3$ .
At $x = 3$ : $\ln4 + \frac12$ .
At $x = 1$ : $\ln2 + 1$ .
Subtract: $\ln4 - \ln2 - \frac12 = \ln2 - \frac12$
$\approx 0.6931 - 0.5$
$= 0.193$ .