Answer
\[ = \frac{{4\sqrt {10} - 4}}{3}\]
Work Step by Step
\[\begin{gathered}
\int_0^3 {\frac{{{v^2} + 1}}{{\sqrt {{v^3} + 3v + 4} }}} \,\,dv \hfill \\
\hfill \\
set\,\,u = {v^3} + 3v + 4\,\,\,\,then\,\,\,du = 3\,\left( {{v^2} + 1} \right)dv \hfill \\
\hfill \\
v = 0\,\,\,\,implies\,\,\,u = 4 \hfill \\
v = 3\,\,\,\,\,implies\,\,\,u = 40 \hfill \\
\hfill \\
apply\,\,the\,\,\,substitution \hfill \\
\hfill \\
= \frac{1}{3}\int_4^{40} {\frac{{du}}{{\sqrt u }}} \hfill \\
\hfill \\
integrate \hfill \\
\hfill \\
= \frac{2}{3}\,\,\,\left[ {\sqrt u } \right]_4^{40} \hfill \\
\hfill \\
Fundamental\,\,theorem \hfill \\
\hfill \\
= \frac{2}{3}\,\left( {\sqrt {40} - \sqrt 4 } \right) \hfill \\
\hfill \\
Simplify \hfill \\
\hfill \\
= \frac{{4\sqrt {10} - 4}}{3} \hfill \\
\end{gathered} \]
