Explanation

(c)
$I = \int_{- 1}^{1}{\frac{\cos{h\ x}}{1 + e^{2x}}\ dx} = \int_{- 1}^{1}{\frac{e^{x} + e^{- x}}{2(1 + e^{2x})}\ dx}\ \ \ \ \ \ \left\lbrack \because\cosh x = \frac{e^{x} + e^{- x}}{2} \right\rbrack$
$= \frac{1}{2}\int_{- 1}^{1}{\frac{1 + e^{2x}}{\left( 1 + e^{2x} \right)e^{x}}\ dx} = \frac{1}{2}\int_{- 1}^{1}{e^{- x}\ dx}$
$= - \frac{1}{2}\left\lbrack e^{- x} \right\rbrack_{- 1}^{1} = - \frac{1}{2}\left( e^{- 1} - e^{1} \right) = \frac{e^{2} - 1}{2e}$