Lore For Math

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

${\displaystyle 1}$

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

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

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

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

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

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