The e^3 is a constant, so bring it outside the integral sign.
Remember proper grouping symbols. You're in calc now.
\(\displaystyle e^{3}\int\frac{1}{1+e^{x}}dx\)
Let, as you correctly stated, \(\displaystyle u=e^{x}, \;\ x=ln(u), \;\ dx=\frac{1}{u}du\)
Then, we get:
\(\displaystyle e^{3}\int\frac{1}{u+u^{2}}du\)
Expand:
\(\displaystyle e^{3}\left[\int\frac{1}{u}-\int\frac{1}{1+u}\right]du\)
Intergate and resub
\(\displaystyle e^{3}\left[ln(u)-ln(1+u)\right]\)
\(\displaystyle e^{3}\left[ln(e^{x})-ln(1+e^{x})\right]\)
\(\displaystyle e^{3}\left(x-ln(e^{x}+1)\right)\)
