Explanation

(d)
Let $I = \int_{}^{}{\frac{e^{x}1 + nx^{n - 1} - x^{2n}}{(1 - x^{n})\sqrt{1 - x^{2n}}}dx}$
$= \int_{}^{}{e^{x}\left\{ \frac{\sqrt{1 - x^{2n}}}{1 - x^{n}} + \frac{nx^{n - 1}}{(1 - x^{n})\sqrt{1 - x^{2n}}} \right\}}dx$
Let $f(x) = \frac{\sqrt{1 - x^{2n}}}{(1 - x^{n})}$
$\Rightarrow f^{'}(x) = \frac{\frac{\left( 1 - x^{n} \right)( - 2n\ x^{2n - 1})}{2\sqrt{1 - x^{2n}}} + \sqrt{1 - x^{2n}}(nx^{n - 1})}{\left( 1 - x^{n} \right)^{2}}$
$\Rightarrow f^{'}(x) = \frac{\left( 1 - x^{2n} \right)\left( nx^{n - 1} \right) - n\left( 1 - x^{n} \right)x^{2n - 1}}{\left( 1 - x^{n} \right)^{2}\sqrt{1 - x^{2n}}}$
$= \frac{n\left( 1 - x^{n} \right)x^{n - 1}\{\left( 1 + x^{n} \right) - x^{n}\}}{\left( 1 - x^{n} \right)^{2}\sqrt{1 - x^{2n}}}$
$= \frac{nx^{n - 1}}{(1 - x^{n})\sqrt{1 - x^{2n}}}$
$\therefore I = \frac{e^{x}\sqrt{1 - x^{2n}}}{\left( 1 - x^{n} \right)} + c$