Explanation

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