Answer
$$ - \ln \left| {x - 1} \right| + 3\ln \left| x \right| - \frac{1}{x} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{2{x^2} - 2x - 1}}{{{x^3} - {x^2}}}} dx \cr
& {\text{Factoring the denominator}} \cr
& \int {\frac{{2{x^2} - 2x - 1}}{{{x^2}\left( {x - 1} \right)}}} dx \cr
& \cr
& {\text{The partial fraction decomposition of the integrand is}} \cr
& \frac{{2{x^2} - 2x - 1}}{{{x^2}\left( {x - 1} \right)}} = \frac{A}{{x - 1}} + \frac{B}{x} + \frac{C}{{{x^2}}} \cr
& {\text{Multiplying by }}{x^2}\left( {x - 1} \right){\text{ yields}} \cr
& 2{x^2} - 2x - 1 = A{x^2} + Bx\left( {x - 1} \right) + C\left( {x - 1} \right) \cr
& 2{x^2} - 2x - 1 = A{x^2} + B{x^2} - Bx + Cx - C \cr
& {\text{Collecting like powers of }}x{\text{, this becomes}} \cr
& 2{x^2} - 2x - 1 = \left( {A{x^2} + B{x^2}} \right) + \left( { - Bx + Cx} \right) - C \cr
& {\text{Equating corresponding coefficients yields the following system }} \cr
& {\text{of linear equations}} \cr
& \,\,\,\,\,A + B = 2 \cr
& - B + C = - 2 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,C = 1 \cr
& {\text{Solving the system of linear equations we obtain}} \cr
& \,\,\,A = - 1,\,\,\,B = 3,\,\,\,C = 1 \cr
& \cr
& {\text{Then}}{\text{, the integrand can be written as}} \cr
& \frac{{2{x^2} - 2x - 1}}{{{x^2}\left( {x - 1} \right)}} = - \frac{1}{{x - 1}} + \frac{3}{x} + \frac{1}{{{x^2}}} \cr
& \int {\frac{{2{x^2} - 2x - 1}}{{{x^2}\left( {x - 1} \right)}}} dx = \int {\left( { - \frac{1}{{x - 1}} + \frac{3}{x} + \frac{1}{{{x^2}}}} \right)} dx \cr
& {\text{Integrating}} \cr
& = - \ln \left| {x - 1} \right| + 3\ln \left| x \right| - \frac{1}{x} + C \cr} $$
