Answer
$\displaystyle \frac{31}{30}$
Work Step by Step
$\displaystyle \int_{0}^{1}\int_{0}^{1}v(u+v^{2})^{4}dudv$ = $\displaystyle \int_{0}^{1}[\int_{0}^{1}v(u+v^{2})^{4}du]dv$ = $\left[\begin{array}{l}
t=u+v^{2}\\
dt=du
\end{array}\right]$ =
... treat v as a constant in the inner integral.
$\displaystyle \int_{0}^{1}[v\int_{v^{2}}^{v^{2}+1}t^{4}dt]dv$
$=\displaystyle \int_{0}^{1}[\frac{vt^{5}}{5}]_{t=v^{2}}^{t=v^{2}+1}dv$
$=\displaystyle \frac{1}{5}\int_{0}^{1}v [(v^{2}+1)^{5}-(v^{2})^{5}]dv$
$=\displaystyle \frac{1}{5}\int_{0}^{1}[v(1+v^{2})^{5}-v^{11}]dv$
$=\displaystyle \frac{1}{5}[\int_{0}^{1}v(1+v^{2})^{5}dv-\int_{0}^{1}v^{11}dv]$
... for the first integral: $\left[\begin{array}{ll}
t=1+v^{2} & \\
dt=2vdv & \\
v=0\Rightarrow t=1, & v=1\Rightarrow t=2
\end{array}\right]$
$=\displaystyle \frac{1}{5}\{\int_{1}^{2}\frac{1}{2}t^{5}dt-[\frac{v^{12}}{12}]_{0}^{1}\}$
$=\displaystyle \frac{1}{5}\{\frac{1}{2} [ \displaystyle \frac{t^{6}}{6}]_{1}^{2}-\frac{1-0}{12}\}$
$=\displaystyle \frac{1}{5}\{\frac{64-1}{12}-\frac{1}{12}\}$
$=\displaystyle \frac{62}{5\cdot 12}$
$=\displaystyle \frac{31}{30}$
