Answer
$$\frac{1}{4}\ln \left| {\frac{{x - 2}}{x}} \right| + \frac{1}{{2x}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{{x^3} - 2{x^2}}}} \cr
& {\text{factor}} \cr
& \int {\frac{{dx}}{{{x^2}\left( {x - 2} \right)}}} \cr
& {\text{partial fractions}} \cr
& \frac{1}{{{x^2}\left( {x - 2} \right)}} = \frac{A}{x} + \frac{B}{{{x^2}}} + \frac{C}{{x - 2}} \cr
& 1 = Ax\left( {x - 2} \right) + B\left( {x - 2} \right) + C{x^2} \cr
& {\text{multiplying}} \cr
& 1 = A{x^2} - 2Ax + Bx - 2B + C{x^2} \cr
& 1 = \left( {A{x^2} + C{x^2}} \right) + \left( { - 2Ax + Bx} \right) - 2B \cr
& {\text{by equating the coefficients}} \cr
& {x^2}:{\text{ }}A + C = 0 \cr
& x:{\text{ }} - {\text{2}}A + B = 0 \cr
& {x^0}:{\text{ }} - 2B = 1 \cr
& {\text{Solving these equations}} \cr
& B = - 1/2 \cr
& A = - 1/4 \cr
& C = 1/4 \cr
& {\text{substituting constants}} \cr
& \frac{A}{x} + \frac{B}{{{x^2}}} + \frac{C}{{x - 2}} = \frac{{ - 1/4}}{x} + \frac{{ - 1/2}}{{{x^2}}} + \frac{{1/4}}{{x - 2}} \cr
& \int {\frac{{dx}}{{{x^3} - 2{x^2}}}} = \int {\left( {\frac{{ - 1/4}}{x} + \frac{{ - 1/2}}{{{x^2}}} + \frac{{1/4}}{{x - 2}}} \right)dx} \cr
& {\text{integrating}} \cr
& = - \frac{1}{4}\ln \left| x \right| + \frac{1}{{2x}} + \frac{1}{4}\ln \left| {x - 2} \right| + C \cr
& = \frac{1}{4}\ln \left| {\frac{{x - 2}}{x}} \right| + \frac{1}{{2x}} + C \cr} $$
