Answer
$$\ln \left( 5 \right)$$
Work Step by Step
$$\eqalign{
& \int_0^2 {\int_0^{2x} {\frac{1}{{{x^2} + 1}}} dydx} \cr
& \int_0^2 {\left[ {\int_0^{2x} {\frac{1}{{{x^2} + 1}}} dy} \right]dx} \cr
& \int_0^2 {\left( {\frac{1}{{{x^2} + 1}}} \right)\left[ {\int_0^{2x} {dy} } \right]dx} \cr
& {\text{Evaluate inner integral}} \cr
& \int_0^{2x} {dy} = \left[ y \right]_{y = 0}^{y = 2x} \cr
& = 2x - 0 \cr
& = 2x \cr
& {\text{Therefore,}} \cr
& \int_0^2 {\left( {\frac{1}{{{x^2} + 1}}} \right)\left[ {\int_0^{2x} {dy} } \right]dx} = \int_0^2 {\left( {\frac{1}{{{x^2} + 1}}} \right)\left[ {2x} \right]dx} \cr
& = \int_0^2 {\frac{{2x}}{{{x^2} + 1}}dx} \cr
& {\text{Integrating}} \cr
& {\text{ = }}\left[ {\ln \left( {{x^2} + 1} \right)} \right]_0^2 \cr
& = \ln \left( {{2^2} + 1} \right) - \ln \left( {{0^2} + 1} \right) \cr
& = \ln \left( 5 \right) \cr} $$
