Explanation

(a)
Put $x + 1 = t^{2}\ \Longrightarrow \ dx = 2t\ dt$
At $x = 8,\ t = 3\ and\ x = 15,\ t = 4$
$\therefore I = \int_{3}^{4}\frac{2t\ dt}{\left( t^{2} - 1 - 3 \right)t}$
$= \int_{3}^{4}\frac{2dt}{(t^{2} - 4)} = 2.\frac{1}{4}\left\lbrack \log\frac{t - 2}{t + 2} \right\rbrack_{3}^{4}$
$= \frac{1}{2}\left\lbrack \log{\frac{1}{3} - \log\frac{1}{5}} \right\rbrack = \frac{1}{2}\log\frac{5}{3}$