Lore For Math

${\displaystyle \underset{‾}{\frac{1}{x(ax+b)}=\frac{1}{b}(\frac{1}{x}-\frac{a}{ax+b})\mathrm{.}}}$

${\displaystyle \frac{1}{x(ax+b)}}$

${\displaystyle =\frac{a}{b}\frac{b}{ax(ax+b)}}$

${\displaystyle =\frac{a}{b}\frac{(ax+b)-ax}{ax(ax+b)}}$

${\displaystyle =\frac{a}{b}\left(\frac{1}{ax}-\frac{1}{ax+b}\right)}$

${\displaystyle =\frac{1}{b}(\frac{1}{x}-\frac{a}{ax+b})\mathrm{.}}$