Answer
$$\frac{{{x^2}}}{2} + x + 2\ln \left| {x - 1} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^2} + 1}}{{x - 1}}} dx \cr
& {\text{The integrand is an improper rational function since the numerator }} \cr
& {\text{has degree 2 and the denominator has degree 1}}{\text{. Thus}}{\text{, using the long division}} \cr
& \,\,\,\,\,\,\frac{{{x^2} + 1}}{{x - 1}} = x + 1 + \frac{2}{{x - 1}} \cr
& {\text{The integrand can be expressed as}} \cr
& \int {\left( {x + 1 + \frac{2}{{x - 1}}} \right)} dx \cr
& {\text{Integrating}} \cr
& = \frac{{{x^2}}}{2} + x + 2\ln \left| {x - 1} \right| + C \cr} $$
