Question:
$\displaystyle \int \frac{dx}{1+e^x} $
How to solve this ..?
Please answer in steps thank you
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.
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:
$\displaystyle \int \frac{du}{u(u-1)} = \int \frac{du}{u^2-u}$