$\displaystyle \int \frac{dx}{\sqrt{1+e^{x}}}$

$\displaystyle u^{2}=1+e^{x} $

[tex] 2u du=e^{x} dx [tex]

I finally get to

$\displaystyle \frac{2}{3} (\int \frac{1}{u-1}du+ \int \frac{u+1}{u^{2}+u+1}du) $

so if this is right

can somebody help me with hte second part of the last integral

and if it's wrong somekind of solution