Calculate the double integral. \int\int\frac{2(1+x^2)}{1+y^2}dA

Step 1
$\text{It is given that }R=\{(x,y)|0\le x\le 2,0\le y\le 1\}.\text{We have to find}$
${\displaystyle \int {\int }_R\frac{2(1+x^2)}{1+y^2}dA}$
Notice that
${\displaystyle \int {\int }_R\frac{2(1+x^2)}{1+y^2}dA={\int }_{\{y=0\}}^1{\int }_{\{x=0\}}^1\frac{2(1+x^2)}{1+y^2}dxdy}$
$$\begin{gathered} ={\int }_0^1[\frac{2}{1+y^2}\cdot {\int }_0^{2}1+x^{2}dx]dy \\ ={\int }_0^1\left[\frac{2}{1+y^2}(2+\frac{8}{3})\right]dy \end{gathered}$$
${\displaystyle ={\int }_0^1\left(\frac{28}{3(1+y^2)}\right)dy}$
${\displaystyle =\frac{28}{3}\cdot {\int }_0^1\frac{1}{1+y^2}dy}$
${\displaystyle =\frac{28}{3}{\left[\arctan \left(y\right)\right]}_0^1[\because \int \frac{1}{1+y^2}dy=\arctan \left(y\right)]}$
${\displaystyle y=\frac{28}{3}\cdot \frac{\pi }{4}}$
${\displaystyle =\frac{7\pi }{3}}$
Result
${\displaystyle {\int }_0^1{\int }_0^2\frac{2(1+x^2)}{1+y^2}dxdy=\frac{7\pi }{3}}$