Answer
\[ = \frac{1}{{16}}\ln \,\left( {\frac{{{v^2}}}{{{v^2} + 8}}} \right) + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{dv}}{{v\,\left( {{v^2} + 8} \right)}}} \hfill \\
\hfill \\
using\,\,formula\,\,85 \hfill \\
\hfill \\
\int_{}^{} {\frac{{dx}}{{x\,\left( {{a^2} + {x^2}} \right)}}} = \frac{1}{{2{a^2}}}\ln \,\left( {\frac{{{x^2}}}{{{a^2} + {x^2}}}} \right) + C \hfill \\
\hfill \\
with\,\,\,{a^2} = 8 \hfill \\
\hfill \\
then \hfill \\
\hfill \\
\int_{}^{} {\frac{{dv}}{{v\,\left( {{v^2} + 8} \right)}}} = \,\frac{1}{{2\,{{\left( 8 \right)}^2}}}\ln \,\left( {\frac{{{v^2}}}{{{v^2} + 8}}} \right) + C \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
= \frac{1}{{16}}\ln \,\left( {\frac{{{v^2}}}{{{v^2} + 8}}} \right) + C \hfill \\
\end{gathered} \]