Answer
$x + \frac{{{x^2}}}{2} + \ln \left| {x - 1} \right| + C$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^2}}}{{x - 1}}} dx \cr
& {\text{The integrand is an improper rational function}}{\text{, then the first }} \cr
& {\text{step is apply the long division:}} \cr
& \frac{{{x^2}}}{{x - 1}} = 1 + x + \frac{1}{{x - 1}} \cr
& {\text{Therefore}}{\text{,}} \cr
& \int {\frac{{{x^2}}}{{x - 1}}} dx = \int {\left( {1 + x + \frac{1}{{x - 1}}} \right)} dx \cr
& = \int {dx} + \int x dx + \int {\frac{1}{{x - 1}}} dx \cr
& {\text{Integrating}} \cr
& = x + \frac{{{x^2}}}{2} + \ln \left| {x - 1} \right| + C \cr} $$
