# Integrals involving exponential

• Feb 1st 2010, 03:38 AM
mj.alawami
Integrals involving exponential
Question:
$\displaystyle \int \frac{dx}{1+e^x}$

How to solve this ..?

• Feb 1st 2010, 03:41 AM
e^(i*pi)
Quote:

Originally Posted by mj.alawami
Question:
$\displaystyle \int \frac{dx}{1+e^x}$

How to solve this ..?

$\displaystyle u = 1+e^x$

$\displaystyle du = e^x dx$

$\displaystyle dx = (u-1)\, du$

$\displaystyle \int \frac{(u-1)}{u} \, du = \int 1\,du - \int \frac{1}{u}\, du$
• Feb 1st 2010, 03:54 AM
tarnishd
Try 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 AM
mj.alawami
Quote:

Originally Posted by e^(i*pi)
$\displaystyle u = 1+e^x$

$\displaystyle du = e^x dx$

$\displaystyle dx = (u-1)\, du$

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

I still didnt understad , Sorry
• Feb 1st 2010, 04:47 AM
e^(i*pi)
Quote:

Originally Posted by mj.alawami
I still didnt understad , Sorry

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:

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