Answer
$$\frac{{{x^2}}}{2} - x + \ln \left| {{x^2} + x - 2} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^3} - x - 3}}{{{x^2} + x - 2}}} dx \cr
& {\text{The integrand is not a proper function, so by long division}} \cr
& \frac{{{x^3} - x - 3}}{{{x^2} + x - 2}} = x - 1 + \frac{{2x + 1}}{{{x^2} + x - 2}} \cr
& {\text{Then,}} \cr
& \int {\frac{{{x^2} + 3x - 4}}{{{x^3} - 4{x^2} + 4x}}} dx = \int {\left( {x - 1 + \frac{{2x + 1}}{{{x^2} + x - 2}}} \right)} dx \cr
& {\text{Integrating}} \cr
& = \int x dx - \int {dx} + \int {\frac{{\overbrace {\left( {2x + 1} \right)dx}^{du}}}{{\underbrace {{x^2} + x - 2}_u}}} \cr
& = \frac{{{x^2}}}{2} - x + \ln \left| {{x^2} + x - 2} \right| + C \cr} $$
