Lore For Math

${\displaystyle \underset{‾}{\frac{1}{x{(ax+b)}^2}=\frac{1}{b{x}^2}-\frac{1}{b^2(ax+b)}-\frac{a}{b^{2}x}-\frac{a}{b^2{(ax+b)}^2}\mathrm{.}}}$

${\displaystyle \frac{1}{x{(ax+b)}^2}}$

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

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

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

${\displaystyle =\frac{1}{b}\left(\frac{1}{x(ax+b)}-\frac{a}{{(ax+b)}^2}\right)}$

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

${\displaystyle =\frac{1}{b}\left(\frac{1}{b}(\frac{1}{x}-\frac{a}{ax+b})-\frac{a}{{(ax+b)}^2}\right)}$

${\displaystyle =\frac{1}{b^{2}x}-\frac{a}{b(ax+b)}-\frac{a}{b{(ax+b)}^2}\mathrm{.}}$