Question

Integrate: $\int \frac{x^4-x^3+8x-8}{x^2-2x+4}dx$

Answer 1

$I=\int \frac{(x^3+8)(x-1)}{x^2-2x+4}dx=\int \frac{(x^3+2^3)(x-1)}{x^2-2x+4}dx\because a^3+b^3=(a+b)(a^2-ab+b^2)=\int \frac{(x+2)(x^2-2x+4)(x-1)}{(x^2-2x+4)}dx=\int (x+2)(x-1)dx=\int (x^2-x+2x-2)dx=\int (x^2+x-2)dx=\int x^{2}dx+\int xdx-2\int dx=\frac{x^3}{3}+\frac{x^2}{2}-2x+C$