Answer
$$\int\frac{e^u}{(1-e^u)^2}du=\frac{1}{1-e^u}+C$$
Work Step by Step
$$A=\int\frac{e^u}{(1-e^u)^2}du$$ $$A=\int e^u(1-e^u)^{-2}du$$
Let $r=1-e^u$.
Then $dr=-e^udu$, so $e^udu=-dr$
Substitute into $A$, we have $$A=-\int r^{-2}dr$$ $$A=-\frac{r^{-1}}{-1}+C$$ $$A=\frac{1}{r}+C$$ $$A=\frac{1}{1-e^u}+C$$