$I=\iint_{R} \frac{1+x^{2}}{1+y^{2}} d A, \operatorname{com} R=\{(x, y) \mid 0 \leq x \leq 2,0 \leq y \leq 1\}$

Step 1 ：First, we set up the double integral: \(I = \int_{0}^{1} \int_{0}^{2} \frac{1+x^2}{1+y^2} dx dy\)

Step 2 ：Next, we integrate with respect to x: \(I = \int_{0}^{1} \left[ x + \frac{x^3}{3} \right]_0^2 dy\)

Step 3 ：Plug in the limits of integration for x: \(I = \int_{0}^{1} (2 + \frac{8}{3}) dy\)

Step 4 ：Simplify the expression inside the integral: \(I = \int_{0}^{1} \frac{14}{3} dy\)

Step 5 ：Integrate with respect to y: \(I = \left[ \frac{14}{3}y \right]_0^1\)

Step 6 ：Plug in the limits of integration for y: \(I = \frac{14}{3}(1) - \frac{14}{3}(0)\)

Step 7 ：Simplify the expression: \(I = \frac{14}{3}\)

Step 8 ：\(\boxed{\frac{14}{3}}\) is the final answer.