Answer
$$\frac{3}{2}\ln \left| {x - 1} \right| - \frac{1}{4}\ln \left( {{x^2} + 1} \right) + \frac{3}{2}{\tan ^{ - 1}}x + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^2} + 2x}}{{{x^3} - {x^2} + x - 1}}} dx \cr
& = \int {\frac{{{x^2} + 2x}}{{{x^2}\left( {x - 1} \right) + x - 1}}} dx \cr
& = \int {\frac{{{x^2} + 2x}}{{\left( {x - 1} \right)\left( {{x^2} + 1} \right)}}} dx \cr
& {\text{Decomposing the integrand into partial fractions}} \cr
& \frac{{{x^2} + 2x}}{{\left( {x - 1} \right)\left( {{x^2} + 1} \right)}} = \frac{A}{{x - 1}} + \frac{{Bx + C}}{{{x^2} + 1}} \cr
& {x^2} + 2x = A\left( {{x^2} + 1} \right) + \left( {Bx + C} \right)\left( {x - 1} \right) \cr
& {x^2} + 2x = A{x^2} + A + B{x^2} - Bx + Cx - C \cr
& {x^2} + 2x = \left( {A{x^2} + B{x^2}} \right) + \left( { - Bx + Cx} \right) + A - C \cr
& {\text{Equating coefficients}} \cr
& A + B = 1 \cr
& - B + C = 2 \cr
& A - C = 0 \cr
& {\text{Solving the system of equations by using a calculator}} \cr
& A = \frac{3}{2},{\text{ }}B = - \frac{1}{2},{\text{ }}C = \frac{3}{2} \cr
& \cr
& \frac{{{x^2} + 2x}}{{\left( {x - 1} \right)\left( {{x^2} + 1} \right)}} = \frac{A}{{x - 1}} + \frac{{Bx + C}}{{{x^2} + 1}} \cr
& \frac{{{x^2} + 2x}}{{\left( {x - 1} \right)\left( {{x^2} + 1} \right)}} = \frac{{3/2}}{{x - 1}} + \frac{{\left( { - 1/2} \right)x + 3/2}}{{{x^2} + 1}} \cr
& \int {\frac{{{x^2} + 2x}}{{\left( {x - 1} \right)\left( {{x^2} + 1} \right)}}} dx = \int {\left( {\frac{{3/2}}{{x - 1}} + \frac{{\left( { - 1/2} \right)x + 3/2}}{{{x^2} + 1}}} \right)} dx \cr
& = \frac{3}{2}\int {\frac{1}{{x - 1}}} dx - \frac{1}{2}\int {\frac{x}{{{x^2} + 1}}} dx + \frac{3}{2}\int {\frac{1}{{{x^2} + 1}}} dx \cr
& {\text{Integrating}} \cr
& = \frac{3}{2}\ln \left| {x - 1} \right| - \frac{1}{4}\ln \left( {{x^2} + 1} \right) + \frac{3}{2}{\tan ^{ - 1}}x + C \cr} $$
