Answer
\[\frac{1}{{15}}\,{\left( {5{r^2} + 2} \right)^{3/2}} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} r \sqrt {5{r^2} + 2} dr \hfill \\
Let\,\,\,u = 5{r^2} + 2 \hfill \\
So\,\,\,that\,\,du = 10rdr\,\,\,Then \hfill \\
\int_{}^{} r \sqrt {5{r^2} + 2} dr = \frac{1}{{10}}\int_{}^{} {10r\sqrt {5{r^2} + 2} dr} \hfill \\
= \frac{1}{{10}}\int_{}^{} {\sqrt u du} \hfill \\
\frac{1}{{10}}\int_{}^{} {{u^{1/2}}du} \hfill \\
Power\,\,rule \hfill \\
\frac{1}{{10}}\,\left( {\frac{{{u^{3/2}}}}{{3/2}}} \right) + C \hfill \\
\frac{1}{{15}}{u^{3/2}} + C \hfill \\
Substituting\,\,5{r^2} + 2\,\,for\,\,u\,\,gives \hfill \\
\frac{1}{{15}}\,{\left( {5{r^2} + 2} \right)^{3/2}} + C \hfill \\
\end{gathered} \]
