Answer
$\dfrac{x^2}{2}+\dfrac{4}{3} \ln |x+2|+\dfrac{2}{3} \ln |x-1|+C$
Work Step by Step
Apply the integration by parts formula as follows: $\int a'(x) b(x)=a(x) b(x)-\int a(x) b'(x)dx$
Re-write the integral into partial fractions.
$\int \dfrac{x^3+x^2}{x^2+x-2}dx=\dfrac{(x^3+x^2-2x+2x) dx}{x^2+x-2}$
Now, $\int \dfrac{x^3+x^2-2x}{x^2+x-2} dx+ \int \dfrac{2x}{x^2+x-2}dx=\int x \space dx+\int \dfrac{\frac{4}{3}}{x+2}+\dfrac{\dfrac{2}{3}}{x-1} dx \\=\dfrac{x^2}{2}+\dfrac{4}{3} \ln |x+2|+\dfrac{2}{3} \ln |x-1|+C$
