Explanation

(a)
$\int_{}^{}{\left( \ \frac{x + 2}{x + 4}\ \right)^{2}e^{x}\ dx} = \int_{}^{}{e^{x}\left\lbrack \frac{x^{2} + 4x + 4}{(x + 4)^{2}} \right\rbrack dx}$
$= \int_{}^{}{e^{x}\left\lbrack \frac{x(x + 4)}{(x + 4)^{2}} + \frac{4}{(x + 4)^{2}} \right\rbrack dx}$
$= \int_{}^{}{e^{x}\left\lbrack \frac{x}{x + 4} + \frac{4}{(x + 4)^{2}} \right\rbrack dx}$
$= \int_{}^{}{\frac{e^{x}\ x}{(x + 4)}dx + \int_{}^{}{\frac{4e^{x}}{(x + 4)^{2}}dx}}$
$= e^{x}\left( \frac{x}{x + 4}\ \right) - \int_{}^{}{\frac{4e^{x}}{(x + 4)^{2}}dx + \int_{}^{}{\frac{4e^{x}}{(x + 4)^{2}}dx}}$
$= \frac{xe^{x}\ }{(x + 4)} + c$