## Calculus with Applications (10th Edition)

$= -\dfrac{1}{3} (x^2+2x-4)^{-3}+C$
In order to integrate this function, we have to use an «u» substitution. Choose an u, derivate it: $u = x^2+2x-4$ $du = 2x+2$ Then substitute: $= \int \dfrac{du}{u^4} = \int u^{-4}$ Then integrate $= -\dfrac{1}{3} u^{-3}+C$ And undo the substitution: $= \boxed{-\dfrac{1}{3} (x^2+2x-4)^{-3}+C}$