What is $\int \frac {e^x} {e^{2x} + 2e^x + 1} \, dx.$?

Question

What is $$\int \frac {e^x} {e^{2x} + 2e^x + 1} \, dx.$$

Wolfram Alpha gives: no result found in terms of standard mathematical functions.

I get $$\log (e^x+1) + \frac 1 {e^x+1}$$ (proof below). My work seems correct to me, but it's different than $\frac {-1} {e^x+1}$ I've seen elsewhere (without explanation). Am I correct? If not, where is my error?

Let $$u(x) = e^x\\ f(x) = \frac {x} {(x+1)^2}\\ \int f(x) \, dx = \log{(x +1)} \, + \frac 1 {x+1}$$ the last of which is verified here. Then

$$\begin{align*} \int \frac {e^x} {e^{2x} + 2e^x + 1} \, dx &= \int \frac {e^x} {(e^x + 1)^2} \, dx \\ &= \int f(u(x)) \cdot u'(x) \, dx \\ &= \log (e^x+1) + \frac 1 {e^x+1}. \end{align*}$$

but this is different than $\frac {-1} {e^x+1}$.

Answer 1

As mentioned in the comments by @Bruno B, the error is that if you let $u = e^x$, then you have to note that $du = e^xdx$, and so the integral is $$\int\frac{1}{(u+1)^2}du=-\frac{1}{u+1}+C=-\frac{1}{e^x+1}+C$$

Answer 2

As mentioned in the comments by @RyszardSzwarc, one has: \begin{align*} \int\frac{e^{x}}{e^{2x} + 2e^{x} + 1}\mathrm{d}x & = \int\frac{\mathrm{d}(e^{x} + 1)}{(e^{x} + 1)^{2}} = -\frac{1}{1 + e^{x}} + C \end{align*}