$\displaystyle \int\frac{e^{3x}}{e^x+e^{-x}}$
can you do this with substitution somehow?
$\displaystyle \int {\frac{{e^{3x} }}
{{e^x + e^{ - x} }}\,dx} = \int {\frac{{e^{4x} }}
{{e^{2x} + 1}}\,dx} .$
Now this is routine, substitute $\displaystyle u=e^{2x},$
$\displaystyle \int {\frac{{e^{3x} }}
{{e^x + e^{ - x} }}\,dx} = \frac{1}
{2}\int {\frac{u}
{{u + 1}}\,du} = \frac{1}
{2}\left( {\int {du} - \int {\frac{1}
{{u + 1}}\,du} } \right).$
From here the rest is quite straightforward.