Explanation

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