Answer
\[ = -\dfrac{1}{3} (x^2+2x-4)^{-3}+C \]
Work Step by Step
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:
\[ = \boxed{-\dfrac{1}{3} (x^2+2x-4)^{-3}+C} \]
