Answer
$$4$$
Work Step by Step
$$\eqalign{
& \int_1^4 {\int_{ - \sqrt x }^{\sqrt x } {\frac{1}{x}} dydx} \cr
& \int_1^4 {\left[ {\int_{ - \sqrt x }^{\sqrt x } {\frac{1}{x}} dy} \right]dx} \cr
& \int_1^4 {\frac{1}{x}\left[ {\int_{ - \sqrt x }^{\sqrt x } {dy} } \right]dx} \cr
& {\text{Evaluate inner integral}} \cr
& \int_{ - \sqrt x }^{\sqrt x } {dy} = \left[ y \right]_{y = - \sqrt x }^{y = \sqrt x } \cr
& = \sqrt x - \left( { - \sqrt x } \right) \cr
& = 2\sqrt x \cr
& {\text{Therefore,}} \cr
& \int_1^4 {\frac{1}{x}\left[ {\int_{ - \sqrt x }^{\sqrt x } {dy} } \right]dx} = \int_1^4 {\frac{1}{x}\left( {2\sqrt x } \right)dx} \cr
& = 2\int_1^4 {{x^{ - 1/2}}dx} \cr
& {\text{Integrating}} \cr
& {\text{ = 2}}\left[ {\frac{{{x^{1/2}}}}{{1/2}}} \right]_1^4 \cr
& = 4\left[ {\sqrt x } \right]_1^4 \cr
& = 4\left( {\sqrt 4 - \sqrt 1 } \right) \cr
& = 4 \cr} $$
