I was looking around, and stumbled accross the fact that
$$ \int \frac{\mathrm{d}u}{(1+\sqrt{u})(1+u)^2\,} = \frac{1}{2} \kappa \left( 1 - \frac{1}{\sqrt{u}\,} \right) - \frac{1}{4} \log\left( 1 + 2\kappa \right) $$ Where $\kappa = \sqrt{u}/(1+u)$
I know I have seen this integral before, but can not forthe life of me remember where. Are there any clever substitutions one can make, because partial fractions makes this a nightmare.