You can separate that equation as

\(\displaystyle{\frac{{{2}{u}'{u}{''}}}{{{1}+{u}^{{2}}}}}={2}{\frac{{{u}'}}{{{u}}}}\)

where both sides are complete differentials which integrate to

\(\displaystyle{\ln{{\left({1}+{u}'^{{2}}\right)}}}={\ln{{\left({u}^{{2}}\right)}}}+{c}\Rightarrow{1}+{u}'^{{2}}={C}{u}^{{2}}\)

Can you continue?

\(\displaystyle{\frac{{{2}{u}'{u}{''}}}{{{1}+{u}^{{2}}}}}={2}{\frac{{{u}'}}{{{u}}}}\)

where both sides are complete differentials which integrate to

\(\displaystyle{\ln{{\left({1}+{u}'^{{2}}\right)}}}={\ln{{\left({u}^{{2}}\right)}}}+{c}\Rightarrow{1}+{u}'^{{2}}={C}{u}^{{2}}\)

Can you continue?