Answer
$$\int_{0}^{1} \int_{0}^{1} v\left(u+v^{2}\right)^{4} d u d v=\frac{31}{30}$$
Work Step by Step
Given $$\int_{0}^{1} \int_{0}^{1} v\left(u+v^{2}\right)^{4} d u d v$$So we have
\begin{aligned} I&=\int_{0}^{1} \int_{0}^{1} v\left(u+v^{2}\right)^{4} d u d v\\
&=\int_{0}^{1}\left[\frac{v\left(u+v^{2}\right)^{5}}{5}\right]_{0}^{1} d v \\
&= \int_{0}^{1} \frac{v\left(1+v^{2}\right)^{5}}{5}\ dv-\int_{0}^{1} \frac{v\left(0+v^{2}\right)^{5}}{5} d v \\
&= \int_{0}^{1} \frac{v\left(1+v^2\right)^5}{5} \ dv- \int_{0}^{1} \frac{ v^{11}}{5} d v \\
&= \frac{1}{10} \int_{0}^{1}2v\left(1+v^2\right)^5 \ dv- \int_{0}^{1} \frac{ v^{11}}{5} d v \\
&= \frac{1}{60} \left(1+v^2\right)^6|_{0}^{1}- \frac{ v^{12}}{60} |_{0}^{1}\\
&= \frac{1}{60} \left(1+1\right)^6-\frac{1}{60} \left(1 \right)^6- \frac{ 1^{12}}{60}-0\\
&= \frac{2^6}{60} -\frac{1}{60} - \frac{ 1}{60}\\
&= \frac{2^6-1-1}{60} \\
&= \frac{2^6-2}{60} \\
&= \frac{31}{30} \\
\end{aligned}
