Answer
$$\frac{{16 - 8\sqrt 2 }}{3}$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\int_y^{y + 2} {\frac{1}{{\sqrt {x + y} }}} dxdy} \cr
& \int_0^1 {\left[ {\int_y^{y + 2} {\frac{1}{{\sqrt {x + y} }}} dx} \right]dy} \cr
& {\text{Evaluate the inner integral}} \cr
& \int_y^{y + 2} {\frac{1}{{\sqrt {x + y} }}} dx = \int_y^{y + 2} {{{\left( {x + y} \right)}^{ - 1/2}}} dx \cr
& = \left[ {\frac{{{{\left( {x + y} \right)}^{1/2}}}}{{1/2}}} \right]_{x = y}^{x = y + 2} \cr
& = 2\left[ {\sqrt {x + y} } \right]_{x = y}^{x = y + 2} \cr
& = 2\left[ {\sqrt {y + 2 + y} - \sqrt {y + y} } \right] \cr
& = 2\left[ {\sqrt {2y + 2} - \sqrt {2y} } \right] \cr
& = 2\sqrt {2y + 2} - 2\sqrt {2y} \cr
& {\text{Therefore,}} \cr
& \int_0^1 {\left[ {\int_y^{y + 2} {\frac{1}{{\sqrt {x + y} }}} dx} \right]dy} \cr
& = \int_0^1 {\left( {2\sqrt {2y + 2} - 2\sqrt {2y} } \right)} \cr
& {\text{Integrating}} \cr
& = \left[ {\frac{{{{\left( {2y + 2} \right)}^{3/2}}}}{{3/2}} - \frac{{{{\left( {2y} \right)}^{3/2}}}}{{3/2}}} \right]_0^1 \cr
& = \frac{2}{3}\left[ {{{\left( {2y + 2} \right)}^{3/2}} - {{\left( {2y} \right)}^{3/2}}} \right]_0^1 \cr
& = \frac{2}{3}\left[ {{{\left( {2\left( 1 \right) + 2} \right)}^{3/2}} - {{\left( {2\left( 1 \right)} \right)}^{3/2}}} \right] - \frac{2}{3}\left[ {{{\left( {2\left( 0 \right) + 2} \right)}^{3/2}} - {{\left( {2\left( 0 \right)} \right)}^{3/2}}} \right] \cr
& = \frac{2}{3}\left( {{4^{3/2}}} \right) - \frac{2}{3}\left( {{2^{3/2}}} \right) - \frac{2}{3}\left( {{2^{3/2}}} \right) \cr
& = \frac{{16}}{3} - \frac{4}{3}\left( {2\sqrt 2 } \right) \cr
& = \frac{{16 - 8\sqrt 2 }}{3} \cr} $$
