Answer
\[\frac{2}{3}{\left( {u - 1} \right)^{3/2}} + 2\,{\left( {u - 1} \right)^{1/2}} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{u}{{\sqrt {u - 1} }}du} \hfill \\
Let\,\,z = u - 1\,\,,\,\,Then\,\,u = z + 1 \hfill \\
So\,\,that\,\,\,du = dz \hfill \\
Then \hfill \\
\int_{}^{} {\frac{u}{{\sqrt {u - 1} }}du} = \int_{}^{} {\frac{{z + 1}}{{\sqrt z }}dz} \hfill \\
Write\,\,the\,\,integrand\,\,as \hfill \\
\int_{}^{} {\,\left( {\frac{z}{{\sqrt z }} + \frac{1}{{\sqrt z }}} \right)dz} = \int_{}^{} {\,\left( {{z^{1/2}} + {z^{ - 1/2}}} \right)dz} \hfill \\
Use\,\,power\,\,rule \hfill \\
\frac{{{z^{3/2}}}}{{3/2}} + \frac{{{z^{1/2}}}}{{1/2}} + C \hfill \\
\frac{2}{3}{z^{3/2}} + 2{z^{1/2}} + C \hfill \\
Substituting\,\,u = u - 1\,\,gives \hfill \\
\frac{2}{3}{\left( {u - 1} \right)^{3/2}} + 2\,{\left( {u - 1} \right)^{1/2}} + C \hfill \\
\hfill \\
\end{gathered} \]