Lore For Math

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

${\displaystyle 1}$

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

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

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

${\displaystyle cite\int \frac{dx}{x(ax+b)}=-\frac{1}{b}\ln \mid \frac{ax+b}{x}\mid +C,}$

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

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