Explanation

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