Explanation

(c)
Let $I = \int_{\log 2}^{x}{\frac{e^{u}}{e^{u}\left( e^{u} - 1 \right)^{1/2}}du}$
Put $e^{u} - 1 = t^{2} \Rightarrow e^{u}du = 2t\ dt$
$\therefore I = \int_{1}^{\sqrt{e^{x} - 1}}{\frac{2t}{\left( t^{2} + 1 \right)t}dt}$
$= 2\int_{1}^{\sqrt{e^{x} - 1}}\frac{dt}{\left( 1 + t^{2} \right)}$
$= 2\left\lbrack {\tan^{- 1}t}\ \right\rbrack_{1}^{\sqrt{e^{x} - 1}}$
$= 2\left\lbrack \tan^{- 1}{\sqrt{e^{x} - 1} - \frac{\pi}{4}} \right\rbrack = \frac{\pi}{6}$ [given]
$\Rightarrow \tan^{- 1}{\sqrt{e^{x} - 1} = \frac{\pi}{12} + \frac{\pi}{4} = \frac{\pi}{3}}$
$\Rightarrow \sqrt{e^{x} - 1} = \tan{\left( \frac{\pi}{3} \right) = \sqrt{3}}$
$\Rightarrow \ \ e^{x} = 3 + 1 = 4$