Explanation

(d)
Let $I = \int_{}^{}\frac{(e^{x} - e^{- x})}{\left( e^{x} + e^{- x} \right)\log{(\cosh x)}}dx$
Put $\log{\left( \cosh x \right) = t} \Rightarrow \ \frac{1}{\cosh x}.\sinh xdx = dt$
$\Rightarrow \ \ \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}}dx = dt$
$\lbrack\because\ \sinh x = \frac{e^{x} - e^{- x}}{2}$
$and\ \cosh x = \frac{e^{x} + e^{- x}}{2}\rbrack$
$\therefore\ I = \int_{}^{}\frac{1}{t}dt = \log t + c$
$= \log{\left( \log{\cosh x} \right) + c}$