# integration with e^x

• Feb 8th 2009, 12:54 AM
twilightstr
integration with e^x
integrate 1+ e^x/(1-e^x)
• Feb 8th 2009, 01:21 AM
Jhevon
Quote:

Originally Posted by twilightstr
integrate 1+ e^x/(1-e^x)

do you mean $\displaystyle \int \left( 1 + \frac {e^x}{1 - e^x} \right) ~dx$ or $\displaystyle \int \frac {1 + e^x}{1 - e^x}~dx$ ?

in either case, consider the substitution $\displaystyle u = 1 - e^x$. (for the second integral, it may be easier to use this substitution if you multiply by $\displaystyle \frac {e^x}{e^x}$ first)
• Feb 8th 2009, 01:38 AM
mr fantastic
Quote:

Originally Posted by Jhevon
do you mean $\displaystyle \int \left( 1 + \frac {e^x}{1 - e^x} \right) ~dx$ or $\displaystyle \int \frac {1 + e^x}{1 - e^x}~dx$ ?

Either way, make the substitution $\displaystyle u = e^x$.
• Feb 10th 2009, 10:00 PM
twilightstr
i dont know how to integrate (1+u)/(u-u^2)
• Feb 10th 2009, 10:19 PM
Watch this
$\displaystyle u= e^{x}$

$\displaystyle \frac{1+e^x} {1-e^x} = \frac{1+u}{1-u}$

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

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

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

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

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