Explanation

(d)
Let $I = \int_{}^{}{\frac{1 + \tan^{2}x}{1 - \tan^{2}x}\ dx} = \int_{}^{}{\frac{\sec^{2}x}{1 - \tan^{2}x}dx}$
Put $\tan{x = t \Rightarrow \sec^{2}{x\ dx = dt}}$
$\therefore I = \int_{}^{}\frac{dt}{1 - t^{2}}$
$= \frac{1}{2 \times 1}\log{\left\lbrack \frac{1 + t}{1 - t} \right\rbrack + c = \frac{1}{2}\log{\left( \frac{1 + \tan x}{1 - \tan x} \right) + c}}$