Question

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

Answer 1

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

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

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

$=\int (x+2)(x-1)dx=\int x^2+x-2dx$

$=\frac{x^3}{3}+\frac{x^2}{2}-2x+c$