Answer
$\displaystyle \frac{1}{3x}+\frac{2}{9}\ln|\frac{2x-3}{x}|+C$
Work Step by Step
$\displaystyle \int\frac{dx}{2x^{3}-3x^{2}}=\int\frac{dx}{x^{2}(-3+2x)}$
$\displaystyle \color{blue}{50.\quad \int\frac{du}{u^{2}(a+bu)}=-\frac{1}{au}+\frac{b}{a^{2}}\ln\left|\frac{a+bu}{u}\right|+C}$
$=-\displaystyle \frac{1}{-3x}+\frac{2}{(-3)^{2}}\ln|\frac{-3+2x}{x}|+C$
$=\displaystyle \frac{1}{3x}+\frac{2}{9}\ln|\frac{2x-3}{x}|+C$
