Answer
$\dfrac{x^2}{2}+\dfrac{4}{3} \ln |x+2|+\dfrac{2}{3} \ln |x-1|+C$
Work Step by Step
Consider the integral $\int \dfrac{x^3+x^2}{x^2+x-2}dx$
The given integral can be re-written as: $\int \dfrac{x^3+x^2}{x^2+x-2}dx=\dfrac{x^3+x^2-2x+2x}{x^2+x-2}dx$
This implies that $\int \dfrac{x^3+x^2-2x}{x^2+x-2} dx+ \int \dfrac{2x}{x^2+x-2}dx=\int x dx+\int \dfrac{\frac{4}{3}}{x+2}+\dfrac{\frac{2}{3}}{x-1} dx$
We use the formula: $\int x^n dx=\int\dfrac{x^{n+1}}{n+1} dx$
$=\dfrac{x^2}{2}+\dfrac{4}{3} \ln |x+2|+\dfrac{2}{3} \ln |x-1|+C$
