Explanation

(d)
Let $I = \int_{}^{}{\frac{2x^{2} + 3}{\left( x^{2} - 1 \right)(x^{2} + 4)}dx = \int_{}^{}\frac{dx}{\left( x^{2} - 1 \right)} + \int_{}^{}\frac{dx}{(x^{2} + 4)}}$
$\left\lbrack \because\ \frac{2x^{2} + 3}{\left( x^{2} - 1 \right)(x^{2} + 4)} = \frac{1}{x^{2} - 1} + \frac{1}{x^{2} + 4}\ \ \right\rbrack$
$\Rightarrow \ I = \frac{1}{2}\log\left( \frac{x - 1}{x + 1} \right) + \frac{1}{2}\tan^{- 1}{\frac{x}{2} + c}$
But $I = a\log{\left( \frac{x - 1}{x + 1} \right) + b\tan^{- 1}\left( \frac{x}{2} \right)} + c$
$\therefore\ \ \ \ a = \frac{1}{2},\ \ \ \ \ b = \frac{1}{2}$