Explanation

(c)
Let $I = \int_{}^{}{e^{x}\frac{x^{2} + 1}{(x + 1)^{2}}}\ dx$
$= \int_{}^{}{e^{x}\ \frac{(x - 1)}{(x + 1)}}dx + 2\int_{}^{}\frac{e^{x}}{(x + 1)^{2}}dx$
$= \int_{}^{}e^{x}dx - 2\int_{}^{}\frac{e^{x}}{(x + 1)}dx + 2\left\lbrack - \frac{e^{x}}{(x + 1)} - \int_{}^{}{- \frac{e^{x}}{(x + 1)}}\ dx \right\rbrack$
$= e^{x} - \frac{2e^{x}}{x + 1} + c = e^{x}\left( \frac{x - 1}{x + 1} \right) + c$