Answer
$$\int_{1}^{3} \int_{0}^{y} \frac{4}{x^{2}+y^{2}} d x d y =\pi \ln 3$$
Work Step by Step
Given$$\int_{1}^{3} \int_{0}^{y} \frac{4}{x^{2}+y^{2}} d x d y$$
So, we have
\begin{align}
\int_{1}^{3} \int_{0}^{y} \frac{4}{x^{2}+y^{2}} d x d y&=
\int_{1}^{3} \int_{0}^{y} \frac{4}{y^{2}((\frac{x }{y })^2+1)} d x d y
\end{align}
Since $$\int\frac{d u}{1+u^{2}} =\arctan (u)$$
So, we get
\begin{align}
\int_{1}^{3} \int_{0}^{y} \frac{4}{x^{2}+y^{2}} d x d y&
=\int_{1}^{3}\left[\frac{4}{y} \arctan \left(\frac{x}{y}\right)\right]_{0}^{y} d y\\
&=\int_{1}^{3} \frac{4}{y} \left( \arctan(1)-\arctan(0) \right)d y\\
&=\int_{1}^{3} \frac{4}{y}\left(\frac{\pi}{4}\right) d y\\
&=\int_{1}^{3} \frac{\pi}{y} d y\\
&=[\pi \ln y]_{1}^{3} \ \ \ \ \ \left(since \int \frac{1}{y} d y=\ln y \right)\\
&=\pi \left(\ln 3-\ln1\right)\\
&=\pi \ln 3
\end{align}
