Answer
$$\int_{0}^{3} \int_{0}^{\infty} \frac{x^{2}}{1+y^{2}} d y \ d x=\frac{9 \pi}{2}$$
Work Step by Step
Given $$\int_{0}^{3} \int_{0}^{\infty} \frac{x^{2}}{1+y^{2}} d y \ d x $$
So, we get
\begin{align}
\int_{0}^{3} \int_{0}^{\infty} \frac{x^{2}}{1+y^{2}} d y \ d x
&=
\int_{0}^{3} x^{2}\int_{0}^{\infty} \frac{1}{1+y^{2}} d y \ d x\\&=\int_{0}^{3}x^{2}\left[ \arctan y\right]_{0}^{\infty} d x\\
&=\int_{0}^ {3} x^{2}\left( \arctan \infty - \arctan 0\right) d x\\
&=\int_{0}^ {3} x^{2}\left(\frac{\pi}{2}-0\right) d x\\
&=\left[\frac{\pi}{2} \cdot \frac{x^{3}}{3}\right]_{0}^{3}\\
&=\frac{9 \pi}{2}
\end{align}
