## Calculus (3rd Edition)

$$\frac{1}{2(e^{-x}+2)^2}+c$$
Let $u=e^{-x}+2$, and then $du=-e^{-x}dx$, hence we have $$\int \frac{e^{-x}dx}{(e^{-x}+2)^3}=-\int \frac{du}{u^3}=-\frac{1}{-2}u^{-2}+c\\ =\frac{1}{2(e^{-x}+2)^2}+c$$