Lore For Math

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

${\displaystyle 1}$

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

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

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

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

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

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