Answer
$$\displaystyle{\frac{\pi \ln5 }{8}}$$
Work Step by Step
Evaluate $\displaystyle{\iint_\limits{R} \frac{1}{1+x^2+y^2}} \mathrm{d}A$.
$R $ is bounded below by $y=0$ (i.e $x$-axis), bounded above by the line $y=x$ and bounded laterally by the circle $x^2+y^2=4$.
Hence, $\theta$ varies from $ 0 $ to $ \pi/4 $, and $r$ varies from $ 0 $ to $ 2. $
\begin{align}
\displaystyle{\iint_\limits{R} \frac{1}{1+x^2+y^2}} \mathrm{d}A &= \displaystyle{\int_{0}^{\pi/4} \int_{0}^{2} \frac{r}{1+r^2}} \mathrm{d}r \,\mathrm{d}\theta \\
&=\displaystyle{\frac{1}{2}\int_{0}^{\pi/4} \ln(1+r^2)} \bigg\rvert_{0}^{2} \,\mathrm{d}\theta \\
&=\displaystyle{\frac{\ln5}{2} \int_{0}^{\pi/4}\mathrm{d}\theta} \\
&=\displaystyle{\frac{\pi \ln5 }{8}}.
\end{align}
