# integrate (e^-x)/(1+e^-x)

• Sep 3rd 2009, 01:43 PM
jebckr
integrate (e^-x)/(1+e^-x)
The answer is -ln(1+e^-x). I'm not real sure how it got there.
• Sep 3rd 2009, 01:45 PM
Matt Westwood
Are you familiar with $\displaystyle \int \frac {f'(x)}{f(x)} dx = \ln (f(x)) + C$?
• Sep 3rd 2009, 01:48 PM
TKHunny
Have you considered:

u = {whatever is in the denominator}

??
• Sep 3rd 2009, 01:54 PM
jebckr
We havn't been taught that rule Matt.

And "u" substitution doesn't give me an ln. I doubt I'll be tested on this, but still wanted to know. Thanks Matt, the rule is something I need to know I guess.
• Sep 3rd 2009, 01:59 PM
Matt Westwood
Okay then, substitute $\displaystyle u = 1 + e^{-x}$. Get $\displaystyle du$ in terms of $\displaystyle x$ and $\displaystyle dx$ and see what happens.
• Sep 3rd 2009, 02:00 PM
Matt Westwood
Quote:

Originally Posted by jebckr
We havn't been taught that rule Matt.

And "u" substitution doesn't give me an ln. I doubt I'll be tested on this, but still wanted to know. Thanks Matt, the rule is something I need to know I guess.

yes it does, you get $\displaystyle -\int \frac 1 u du$ which is wot?
• Sep 3rd 2009, 02:10 PM
jebckr
ok, thanks. Got it.