To evaluate the integral, we have a substitution.

`int{e^(3x)}/{e^x+1}dx`

`=int{(e^x)^3}/{e^x+1}dx` let `u=e^x` , so `du=e^xdx`

`=int u^2/{u+1}du` now let `v=u+1` so `dv=du`

`=int(v-1)^2/vdv` expand numerator

`=int{v^2-2v+1}/vdv` simplify

`=int(v^2/v-{2v}/v+1/v)dv` simplify further

`=int(v-2+1/v)dv` now integrate with power rule

`=1/2v^2-2v+lnv+K` where K is the constant of integration

`=1/2(u+1)^2-2(u+1)+ln(u+1)+K`

`=1/2u^2-u-1/2+ln(u+1)+K`

`=1/2u^2-u+ln(u+1)+C` where C=K+1 is a constant

`=1/2e^{2x}-e^x+ln(e^x+1)+C`

**The integral evaluates to `1/2e^{2x}-e^x+ln(e^x+1)+C` .**