Question

integrate this

Answer 1

\[\int\limits_{}^{}\frac{ 1 }{ e^u+1}\]

Answer 2

This is a interesting problem!

Answer 3

Tried U-sub, by parts, partial frac noting is working

Answer 4

I'll try my hand at it. Just a second.

Answer 5

I can share my way if no one else has ideas! :)

Answer 6

If you know already, you should :)

Answer 7

Well, initially it does not appear there is anything to u-sub with. But, the derivative of the e^x function is itself. How about multiplying both numerator and denominator by e^u ?
\( \displaystyle \int \frac{1}{e^u + 1} \cdot \frac{e^u}{e^u} \ du = \int \frac{e^u}{e^{2u} + e^u} \ du \)
That puts a nice little e^u du in our numerator!

Answer 8

So if we substitute, say, v = e^u. dv = e^u du.
\( \displaystyle \int \frac{\color{blue}{e^u}}{(\color{red}{e^u})^2 + \color{red}{e^u}} \ \color{blue}{du} = \int \dfrac{1}{v^2 + v} \ dv \)
The new integral should work with partial fraction decomposition!

Answer 9

Another way, which might be more tricky than the above:
We can start with an integral form we know:
\( \displaystyle \int \frac{e^u + 1}{e^u + 1} \ du = \int \frac{e^u}{e^u + 1} \ du + \int \frac{1}{e^u + 1} \ du \)
The left side is really just the integral of 1 du. And the first integral on the right works with u-sub as well, so we can solve that as well. Then we just solve algebraically for the remaining integral.