Answer
\begin{aligned} \iint_{D} \frac{y}{x^5+1}d A=\frac{\ln2}{10}
\end{aligned}
Work Step by Step
Given $$ \iint_{D} \frac{y}{x^5+1}d A$$
$$D=\{(x, y) |0\leq x \leq 1,0\leq y \leq x^2\}$$
So, we have
\begin{aligned} A&=\iint_{D} \frac{y}{x^5+1} d A\\
&=\int_{0}^{1} \int_{0}^{x^2} \frac{y}{x^5+1 }d y d x\\
&=\int_{0}^{1} \frac{1}{x^{5}+1}\left[\frac{y^{2}}{2}\right]_{0}^{x^{2}} d x\\&=\frac{1}{2} \int_{0}^{1} \frac{x^{4}}{x^{5}+1} d x\\
&=\frac{1}{10} \int_{0}^{1} \frac{5 x^{4}}{x^{5}+1} d x\\
&=\frac{1}{10}\left[ \ln(x^5+1)\right]_{0}^{1}\\
&=\frac{1}{10}\left[ \ln2+\ln1)\right] \\
&=\frac{1}{10} \ln2\end{aligned}
