Answer
\[\frac{{{u^{3/2}}}}{{3/2}} + \frac{{{u^{ - 1}}}}{{ - 1}} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\,\left( {\sqrt u + \frac{1}{{{u^2}}}} \right)du} \hfill \\
Write\,\,\sqrt u \,as\,\,{u^{1/2}}\,\,and\,\,\frac{1}{{{u^2}}} = {u^{ - 2}} \hfill \\
Use\,\,the\,\,power\,\,rule\,\,\int_{}^{} {{u^n}du} = \frac{{{u^{n + 1}}}}{{n + 1}} + C \hfill \\
Then \hfill \\
\frac{{{u^{1/2}}}}{{1/2 + 1}} + \frac{{{u^{ - 2 + 1}}}}{{ - 2 + 1}} + C \hfill \\
Simplifying \hfill \\
\frac{{{u^{3/2}}}}{{3/2}} + \frac{{{u^{ - 1}}}}{{ - 1}} + C \hfill \\
\end{gathered} \]
