Answer
$${x^2} + \ln \left| {\frac{{x - 1}}{x}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{2{x^3} - 2{x^2} + 1}}{{{x^2} - x}}} dx \cr
& {\text{Apply long division to the integrand}} \cr
& \frac{{2{x^3} - 2{x^2} + 1}}{{{x^2} - x}} = 2x + \frac{1}{{{x^2} - x}} \cr
& {\text{Then}}{\text{,}} \cr
& \int {\frac{{2{x^3} - 2{x^2} + 1}}{{{x^2} - x}}} dx = \int {\left( {2x + \frac{1}{{{x^2} - x}}} \right)} dx \cr
& = \int {2x} dx + \int {\frac{1}{{{x^2} - x}}} dx \cr
& = {x^2} + \int {\frac{1}{{{x^2} - x}}} dx \cr
& \cr
& {\text{Solve the integral by the method of partial fractions.}} \cr
& {\text{The form of the partial fraction decomposition is}} \cr
& \frac{1}{{{x^2} - x}} = \frac{1}{{x\left( {x - 1} \right)}} = \frac{A}{x} + \frac{B}{{x - 1}} \cr
& {\text{Multiplying by }}{x^2} + 2x{\text{, we have}} \cr
& {\text{1}} = A\left( {x - 1} \right) + Bx \cr
& {\text{if we set }}x = 0, \cr
& {\text{1}} = A\left( { - 1} \right) \cr
& A = - 1 \cr
& {\text{if we set }}x = 1 \cr
& {\text{1}} = A\left( 0 \right) + B\left( 1 \right) \cr
& B = 1 \cr
& {\text{then}} \cr
& \frac{1}{{x\left( {x - 1} \right)}} = \frac{A}{x} + \frac{B}{{x - 1}} = - \frac{1}{x} + \frac{1}{{x - 1}} \cr
& \int {\frac{1}{{{x^2} - x}}} dx = \int {\left( { - \frac{1}{x} + \frac{1}{{x - 1}}} \right)} dx \cr
& \int {\frac{1}{{{x^2} - x}}} dx = \int {\left( { - \frac{1}{x} + \frac{1}{{x - 1}}} \right)} dx \cr
& = - \ln \left| x \right| + \ln \left| {x - 1} \right| + C \cr
& = \ln \left| {\frac{{x - 1}}{x}} \right| + C \cr
& \cr
& {x^2} + \int {\frac{1}{{{x^2} - x}}} dx = {x^2} + \ln \left| {\frac{{x - 1}}{x}} \right| + C \cr} $$
