Find $\displaystyle \int \frac{1+e^x}{1-e^x}$
Thanks!
$\displaystyle \int {\frac{{1 + e^x }}
{{1 - e^x }}\,dx} = \int {\frac{{1 - e^x + 2e^x }}
{{1 - e^x }}\,dx} = \int {dx} - 2\int {\frac{{\left( {1 - e^x } \right)'}}
{{1 - e^x }}\,dx}.$
So $\displaystyle \int {\frac{{1 + e^x }}
{{1 - e^x }}\,dx}=x - 2\ln \left| {1 - e^x } \right| + k.$