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