Answer
$$\iint \limits_{R} \frac{x y^{2}}{x^{2}+1} d A=9 \ln 2$$
Work Step by Step
Given$$\iint \limits_{R} \frac{x y^{2}}{x^{2}+1} d A, \quad R=\{(x, y) | 0 \leqslant x \leqslant 1,-3 \leqslant y \leqslant 3\}$$
So, we have
\begin{aligned}I&= \iint \limits_{R} \frac{x y^{2}}{x^{2}+1} dA \\
&= \int_{-3}^{3} \int_{0}^{1} \frac{x y^{2}}{x^{2}+1} d x \ dy\\
&=\left[\int_{-3}^{3} y^{2} d y\right]\left[\frac{1}{2}\int_{0}^{1} \frac{2x}{x^{2}+1} d x\right]\\
&=\left[ \frac{y^{3}}{3} \right]_{-3}^{3}\left[ \frac{1}{2} \ln [x^{2}+1] d x\right]_{0}^{1}\\
&=\left[ \frac{3^{3}}{3} -\frac{ (-3)^{3}}{3}\right] \left[ \frac{1}{2} \ln [1^{2}+1] - \frac{1}{2} \ln [0^{2}+1] \right] \\
&= \frac{2 (3^{3})}{3} \frac{1}{2} \ln 2\\
&= 9 \ln 2
\end{aligned}
