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## Integral explanation [closed]

Can someone explain me, why result $\int \frac{e^x}{e^x+1}$ is $ln(-e^x-1)$ according to WolframAlpha. It is all clear for me until last step (if you click show steps).

I guess that $ln(e^x+1)$ is solution for all non-complex values of x, and $ln(-e^x-1)$ is solution for complex values of x. Am I right?

I have not had any complex analysis course yet, so I am curious :)

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The functions $\ln(e^x+1)$ and $\ln(-e^x-1)$ differ by a constant, $\ln(-1)$, which of course has no meaning over $\mathbb R$. But in the complex variable case any of the two functions is a perfect antiderivative. – Wadim Zudilin May 16 2010 at 13:44
This is the sort of thing that computer algebra systems do all the time. They have no common sense about what is positive or negative :-( – Robin Chapman May 16 2010 at 14:04
That last step is amusingly superfluous. I'm afraid I'm voting to close the question, since it is somewhat off-topic for this site. You may want to have a look at the FAQ for a list of sites that answer similar questions. – S. Carnahan May 16 2010 at 16:19
Closed. We're not interested in questions about WolframAlpha. – Scott Morrison May 16 2010 at 19:12