Answer
$\displaystyle \frac{\pi}{3}$
Work Step by Step
$\displaystyle \iint_{R}\frac{1+x^{2}}{1+y^{2}}dA=\int_{0}^{1}\int_{0}^{1}\frac{1+x^{2}}{1+y^{2}}dydx$
Let $ g(x)=1+x^{2},\displaystyle \quad h(y)=\frac{1}{1+y^{2}}\quad$ and apply formula (5),
$\displaystyle \iint_{R}g(x)\cdot h(y)dA=\int_{a}^{b}g(x)dx\cdot\int_{c}^{d}h(y)dy$
$=\displaystyle \int_{0}^{1}(1+x^{2})dx\cdot\int_{0}^{1}\frac{1}{1+y^{2}}dy$
$=[x+\displaystyle \frac{x^{3}}{3}]_{0}^{1}\cdot [\tan^{-1}y]_{0}^{1}$
$=[1+\displaystyle \frac{1}{3}-(0+0)](\tan^{-1}1-\tan^{-1}0)$
$=[\displaystyle \frac{4}{3}](\frac{\pi}{4}-0)$
$=\displaystyle \frac{\pi}{3}$
