The answer is -ln(1+e^-x). I'm not real sure how it got there.

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- Sep 3rd 2009, 01:43 PMjebckrintegrate (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 PMMatt Westwood
Are you familiar with $\displaystyle \int \frac {f'(x)}{f(x)} dx = \ln (f(x)) + C$?

- Sep 3rd 2009, 01:48 PMTKHunny
Have you considered:

u = {whatever is in the denominator}

?? - Sep 3rd 2009, 01:54 PMjebckr
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 PMMatt 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 PMMatt Westwood
- Sep 3rd 2009, 02:10 PMjebckr
ok, thanks. Got it.