Lore For Math

${\displaystyle \underset{‾}{\int \frac{x^2}{{(ax+b)}^2}dx=\frac{1}{a^3}(ax+b-2b\ln \mid ax+b\mid -\frac{b^2}{ax+b})+C\mathrm{.}}}$

${\displaystyle 1}$

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

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

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

${\displaystyle =\frac{1}{a^3}\left(\int du-\int \frac{2b}{u}du+\int \frac{b^2}{u^2}du\right)}$

${\displaystyle =\frac{1}{a^3}\left(u-2b\ln \mid u\mid -\frac{b^2}{u}\right)}$

${\displaystyle =\frac{1}{a^3}(ax+b-2b\ln \mid ax+b\mid -\frac{b^2}{ax+b})+C\mathrm{.}}$