Explanation

(b)
$I = \int_{}^{}\frac{x^{2}}{(ax + b)^{2}}dx$
Put $ax + b = t\ \Rightarrow dx = \frac{1}{a}\ dt$
and $x = \left( \ \frac{t - b}{a}\ \right)$
$\therefore I = \frac{1}{a^{3}}\int_{}^{}{\frac{(t - b)^{2}}{t^{2}}dt = \frac{1}{a^{3}}\int_{}^{}\left( 1 + \frac{b^{2}}{t^{2}} - \frac{2b}{t} \right)dt}$
$= \frac{1}{a^{3}}\left( t - \frac{b^{2}}{t} - 2b\log t \right) + c$
$= \frac{1}{a^{3}}\left( ax + b - \frac{b^{2}}{ax + b} - 2b\log{(ax + b)} \right) + c$
$= \frac{2}{a^{2}}\left( x - \frac{b}{a}\log{(ax + b)} \right) - \frac{x^{2}}{a(ax + b)} + c$