Lore For Math

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

${\displaystyle 1}$

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

${\displaystyle cite\frac{1}{x{(ax+b)}^2}=\frac{1}{b{x}^2}-\frac{1}{b^2(ax+b)}-\frac{a}{b^{2}x}-\frac{a}{b^2{(ax+b)}^2},}$

${\displaystyle =\frac{1}{b^2}\left(\int \frac{1}{ax+b}dx-a\int \frac{1}{x}dx-a\int \frac{1}{{(ax+b)}^2}dx\right)}$

${\displaystyle cite\int \frac{a}{{(ax+b)}^2}dx=\frac{1}{ax+b}+C,}$

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

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