Lore For Math

${\displaystyle \underset{‾}{\int \frac{x}{{(ax+b)}^2}dx=\frac{1}{a^2}(\ln \mid \frac{ax+b}{x}\mid +\frac{b}{ax+b})+C\mathrm{.}}}$

${\displaystyle 1}$

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

${\displaystyle letu=ax+b,du=adx,dx=\frac{du}{a},x=\frac{u-b}{a},}$

${\displaystyle =\int \frac{\frac{u-b}{a}}{u^2}\frac{du}{a}}$

${\displaystyle =\frac{1}{a^2}\left(\int \frac{1}{u}du-\int \frac{b}{u^2}du\right)}$

${\displaystyle =\frac{1}{a^2}\left(\ln \mid u\mid +\frac{b}{u}\right)}$

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