Why does $\int \frac{e^x}{(1-e^x)^2}dx = - \ln|1-e^x|+C$?

Question

I worked through $\int \frac{e^x}{(1-e^x)^2}dx$ using u-substitution, but my answer, $(1-e^x)^{-1}+C$ is incorrect. It should be $- \ln|1-e^x|+C$

$$\int \frac{e^x}{(1-e^x)^2}dx$$

Let $u = 1 - e^x$

Then $du = -e^x$ and $dx = \frac{-du}{e^x}$

So $\int \frac{e^x}{u^2} \frac{-1}{e^x}du = -\int \frac{1}{u^2}du = -\int u^{-2}du$

$$-\int u^{-2} du$$

$$=-(\frac{u^{-1}}{-1})+C$$

$$=u^{-1}+C$$

$$=(1-e^x)^{-1}+C$$

What did I do wrong?

Answer 1

The result is false: in fact,

$$\int\frac{e^x}{(1-e^x)^2}dx=\frac1{1-e^x} +C$$

This is: what you did is completely correct. There is not $\;\ln\;$ here.

Answer 2

Your answer is perfectly fine and agrees with what WolframAlpha returns. You should give a source as the source seems to be wrong in its given answer.

Answer 3

Your answer is the correct one, as you can confirm by differentiating both. In general you can always confirm whether you got the right primitive by differentiation.