Question

∫$frac{x^{2}+x-3}{6}dx$

Answer 1

$frac{1}{6}int(x^2+x-3)dx=frac{1}{6}int x^2dx+frac{1}{6}int xdx-frac{1}{6}int 3dx=frac{x^3}{18}+frac{x^2}{12}-frac{x}{2}+C$