Explanation

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