How do I evaluate this integral by substitution?

Question

How do I evaluate $$\int \frac{e^{-2x}}{1+e^{-x}} dx$$

I've tried the substitution u = $e^{-x}$, and the furthest I've gotten is $\int \frac{-u}{1+u} du$ and I don't know how to proceed from there.

Answer 1

You've already done the hard part. Notice now that $$\int \frac{u}{1 + u} du = \int \left(1 - \frac{1}{1 + u}\right) du$$

$$\cdots = u - \log |u + 1| + C.$$

Answer 2

$$\int\frac{-u}{1+u}du=\int\frac{-u\color{red}{-1+1}}{1+u}du=-\int du+\int\frac{du}{1+u}=-u+\ln|1+u|+C$$