Explanation

(d)
Let $I = \int_{}^{}{\frac{x^{2} - 1}{\left( x^{4} + 3x^{2} + 1 \right)\tan^{- 1}\left( \ x + \frac{1}{x} \right)}\ dx}$
$= \int_{}^{}{\frac{\left( \ 1 - \frac{1}{x^{2}} \right)}{\left( \ x^{2} + \frac{1}{x^{2}} + 3\ \right)\tan^{- 1}\left( \ x + \frac{1}{x}\ \right)}dx}$
Put $x + \frac{1}{x} = t$
$\Rightarrow \ \left( \ 1 - \frac{1}{x^{2}} \right)dx = dt$ and $\left( \ x + \frac{1}{x}\ \right)^{2} = t^{2}$
$\Rightarrow x^{2} + \frac{1}{x^{2}} = t^{2} - 2$
$\therefore I = \int_{}^{}{\frac{1}{\left( t^{2} + 1 \right)\tan^{- 1}t}dt}$
$= \log{(\tan^{- 1}{t) + c}}$
$= \log{\left\lbrack \tan^{- 1}\left( \ x + \frac{1}{x}\ \right) \right\rbrack + c}$