Question

Solve $\int \frac{(x^3+8)(x-1)}{x^2-2x+4}dx$

Answer 1

$\int \frac{(x^3+8)(x-1)}{x^2-2x+4}dx=\int \frac{(x+2)(x^2-2x+4)(x-1)}{(x^2-2x+4)}dx$

$=\int (x+2)(x-1)dx=\int (x^2-x+2x-2)dx=\int (x^2+x-2)dx=\frac{x^3}{3}+\frac{x^2}{2}-2x+c$

where $c$ is a constant of integration