Answer
$$\ln \left[ {{x^2}{{\left( {x - 1} \right)}^3}} \right] + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{5x - 2}}{{{x^2} - x}}} dx \cr
& {\text{Decompose the integrand into partial fractions}} \cr
& \frac{{5x - 2}}{{{x^2} - x}} = \frac{{5x - 2}}{{x\left( {x - 1} \right)}} \cr
& \frac{{5x - 2}}{{x\left( {x - 1} \right)}} = \frac{A}{x} + \frac{B}{{x - 1}} \cr
& 5x - 2 = A\left( {x - 1} \right) + Bx \cr
& {\text{For }}x = 0 \cr
& - 2 = A\left( { - 1} \right) \to A = 2 \cr
& {\text{For }}x = 1 \cr
& 5 - 2 = B\left( 1 \right) \to B = 3 \cr
& \frac{{5x - 2}}{{x\left( {x - 1} \right)}} = \frac{2}{x} + \frac{3}{{x - 1}} \cr
& {\text{Therefore}} \cr
& \int {\frac{{5x - 2}}{{{x^2} - x}}} dx = \int {\left( {\frac{2}{x} + \frac{3}{{x - 1}}} \right)} dx \cr
& {\text{Integrating}} \cr
& {\text{ = 2ln}}\left| x \right| + 3\ln \left| {x - 1} \right| + C \cr
& {\text{ = }}\ln {x^2} + \ln {\left( {x - 1} \right)^3} + C \cr
& {\text{ = }}\ln \left[ {{x^2}{{\left( {x - 1} \right)}^3}} \right] + C \cr} $$
