Question

Tfraiz.

Answer 1

\[\int\limits \frac{dx}{1-e^{-x}}\]
Add and subtract e^(-x)
\[\int\limits \frac{1+e^{-x}-e^{-x}}{1-e^{-x}}dx=\int\limits \frac{1-e^{-x}}{1-e^{-x}}dx+\int\limits \frac{e^{-x}}{1-e^{-x}}dx\]
Using a u-sub on the second integral.
u=1-e^-x
du=e^-x dx
So you have:
\[\int\limits dx+\int\limits \frac{du}{u}=x+\ln|u|+C=x+\ln|1-e^{-x}|+C\]

Answer 2

If the latex looks bad refresh the page.

Answer 3

Man i totally forgot about that approach ... I tried multiplying through by 1+e^(-x) and just e^(-x) and all sorts of variations of that

Answer 4

dfoydor do you see anything wrong with this?

Answer 5

No, it all seems fine.

Answer 6

Amistre, can you explain why wolfram's answer is different? http://www.wolframalpha.com/input/?i=integral+of+1%2F(1-e^(-x))dx

Answer 7

Wolfram says an alternative form of the answer is such and such; ill have to review later when I got more time if need be :)

Answer 8

I get ln|e^x-1| where wolfram gets ln|1-e^x| i don't know where I lost the negative. I just did it two separate methods (the +/- method and partial fractions) and I get the same thing :/

Answer 9

I checked their answer, and your answer is equivalent up to the second to last step. Instead of turning ln(e^-x) into -x, they combine logs, giving ln( (e^(-x)-1) / (e^-x)), which is ln( (e^-x * e^x - 1*e^x) / 1), which is ln(1-e^x). Both answers are the same.

Answer 10

Okay. I was wondering, I just realized a manipulation would fix that. :D