Lore For Math

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

${\displaystyle 1}$

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

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

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

${\displaystyle cite\int \frac{a}{ax+b}dx=\ln \mid ax+b\mid +C,}$

${\displaystyle =\frac{1}{b}\left(\ln \mid x\mid -\ln \mid ax+b\mid \right)+C}$

${\displaystyle =-\frac{1}{b}\ln \mid \frac{ax+b}{x}\mid +C\mathrm{.}}$