integration with e^x

**twilightstr** · February 8th 2009, 12:54 AM

integrate 1+ e^x/(1-e^x)

**Jhevon** · February 8th 2009, 01:21 AM

Originally Posted by twilightstr

integrate 1+ e^x/(1-e^x)

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

in either case, consider the substitution $u = 1 - e^x$ . (for the second integral, it may be easier to use this substitution if you multiply by $\frac {e^x}{e^x}$ first)

**mr fantastic** · February 8th 2009, 01:38 AM

Originally Posted by Jhevon

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

Either way, make the substitution $u = e^x$ .

**twilightstr** · February 10th 2009, 10:00 PM

i dont know how to integrate (1+u)/(u-u^2)

**ADARSH** · February 10th 2009, 10:19 PM

Watch this
$ u= e^{x} $

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

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

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

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

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

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

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