Question:

$\displaystyle \int \frac{dx}{1+e^x} $

How to solve this ..?

Please answer in steps thank you

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- Feb 1st 2010, 03:38 AMmj.alawamiIntegrals involving exponential
Question:

$\displaystyle \int \frac{dx}{1+e^x} $

How to solve this ..?

Please answer in steps thank you - Feb 1st 2010, 03:41 AMe^(i*pi)
- Feb 1st 2010, 03:54 AMtarnishdTry this...
Let $\displaystyle e^x$=t

x=ln(t)

dx=(1/t)*dt

I=$\displaystyle \int 1/(1+e^x) dx = \int [(1/(t+1)*(1/t)] dt$

I=$\displaystyle \int$ {(1/t)-[1/(t+1)]} dt

$\displaystyle I=ln(t)-ln(t+1)=ln( e^x)-ln( e^x+1)$

I=ln[ex/( ex+1)]+C

I found this proof on the web but verified it in calculation. Seems right to me, hope this helps. - Feb 1st 2010, 04:33 AMmj.alawami
- Feb 1st 2010, 04:47 AMe^(i*pi)
I used integration by substituion.

Let $\displaystyle u = 1+e^x $ - this was chosen to make the denominator easier

When we differentiate u wrt x we get $\displaystyle du = e^x \, dx$ which is obtained through normal rules.

To be able to substitute our du term into the equation we need to isolate dx: $\displaystyle dx = \frac{du}{e^x}$. However that isn't very helpful so to get that e^x out the way rearrange the original substitution

$\displaystyle e^x = u-1$

$\displaystyle dx = \frac{du}{u-1}$

(that was a mistake I made in my previous post. It should read the above)

Putting this back into the integral for dx:

$\displaystyle \int \frac{du}{u(u-1)} = \int \frac{du}{u^2-u}$