$$\iint_R \frac{1}{1+x^2+y^2} \,dA$$
$$R=\left\{(r,\theta):1\le r\le 2,0\le \theta \le \pi\right\}$$
limits of outer integral are $0$ to $\pi$ and inner integral are $1$ to $2$. I wanted to confirm if i did the problem right.
My answeR: $(1/2)\ln(5/2)\pi$
$$ I = \int_0^\pi \left[ \int_1^2\frac 1{1+r^2}r\,dr \right] \, d\theta = \int_0^\pi \left[ [\log(1+ r^2)/2]_1^2 \right] \, d\theta \\ =\int_0^\pi \frac 12 \log\frac 52 \, d\theta = \frac \pi2 \log\frac 52 $$
Hence your answer is correct.
Let $r=\sqrt{x^2+y^2}$ and $dA=dxdy =rdrd\theta$, then $$\int \int_{R}\frac{dA}{1+x^2+y^2}=\int_{\theta=0}^{\theta=\pi}\int_{r=1}^{r=2}\frac{r drd\theta}{1+r^2}= $$ $$=\int_{\theta =0}^{\theta =\pi}\left(\frac{1}{2} \int_{r=1}^{r=2}\frac{2r dr}{1+r^2}\right)d\theta=\int_{\theta=0}^{\theta=\pi}d\theta \cdot \left.\frac{1}{2}\ln(1+r^2)\right|_{r=1}^{r=2}= $$ $$=\pi\frac{1}{2} (\ln(1+4)-\ln(1+1))=\frac{\pi}{2}\ln\frac{5}{2}.$$
