Explanation

(b)
Let $I = \int_{0}^{1}{\frac{\tan^{- 1}x}{1 + x^{2}}\ dx}$
Put $\tan^{- 1}{x = t \Rightarrow \frac{1}{1 + x^{2}}}dx = dt$
$\therefore\ I = \int_{0}^{\pi/4}{t\ dt = \left\lbrack \frac{t^{2}}{2} \right\rbrack_{0}^{\pi/4}}$
$= \frac{1}{2}\left\lbrack \left( \frac{\pi}{4} \right)^{4} - 0^{2} \right\rbrack = \frac{\pi^{2}}{32}$