Explanation

(d)
Given, $\int_{0}^{p}{\frac{dx}{1 + 4x^{2}} = \frac{\pi}{8}}$
LHS$= \int_{0}^{p}{\frac{dx}{1 + 4x^{2}} = \frac{1}{4}}\int_{0}^{p}\frac{dx}{\frac{1}{4} + x^{2}}$
$= \frac{1}{4} \times \frac{1}{\frac{1}{2}}\left\lbrack \tan^{- 1}\left( \frac{x}{\frac{1}{2}} \right) \right\rbrack_{0}^{p}$
$\Rightarrow \frac{1}{2}\ \left\lbrack \tan^{- 1}{2p} \right\rbrack = \frac{\pi}{8}$
$\Rightarrow \tan^{- 1}{2p} = \frac{\pi}{4}$
$\Rightarrow 2p = 1$
$\Rightarrow p = \frac{1}{2}$