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

Calculate the double integral.
$\int \int \frac{2\left(1+{x}^{2}\right)}{1+{y}^{2}}dA,$
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Step 1

$\int {\int }_{R}\frac{2\left(1+{x}^{2}\right)}{1+{y}^{2}}dA$
Notice that
$\int {\int }_{R}\frac{2\left(1+{x}^{2}\right)}{1+{y}^{2}}dA={\int }_{\left\{y=0\right\}}^{1}{\int }_{\left\{x=0\right\}}^{1}\frac{2\left(1+{x}^{2}\right)}{1+{y}^{2}}dxdy$
$={\int }_{0}^{1}\left[\frac{2}{1+{y}^{2}}\cdot {\int }_{0}^{2}1+{x}^{2}dx\right]dy\phantom{\rule{0ex}{0ex}}={\int }_{0}^{1}\left[\frac{2}{1+{y}^{2}}\left(2+\frac{8}{3}\right)\right]dy$
$={\int }_{0}^{1}\left(\frac{28}{3\left(1+{y}^{2}\right)}\right)dy$
$=\frac{28}{3}\cdot {\int }_{0}^{1}\frac{1}{1+{y}^{2}}dy$

$y=\frac{28}{3}\cdot \frac{\pi }{4}$
$=\frac{7\pi }{3}$
Result
${\int }_{0}^{1}{\int }_{0}^{2}\frac{2\left(1+{x}^{2}\right)}{1+{y}^{2}}dxdy=\frac{7\pi }{3}$