Answer
$$\frac{\pi }{8}$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\int_0^1 {\frac{y}{{1 + {x^2}}}} } dxdy \cr
& = \int_0^1 {\left[ {\int_0^1 {\frac{y}{{1 + {x^2}}}} dx} \right]} dy \cr
& {\text{solve the inner integral}}{\text{, treat }}y{\text{ as a constant}} \cr
& = \int_0^1 {\frac{y}{{1 + {x^2}}}} dx \cr
& = y\int_0^1 {\frac{1}{{1 + {x^2}}}} dx \cr
& = y\left[ {\arctan x} \right]_0^1 \cr
& {\text{evaluating the limits for }}x \cr
& = y\left[ {\arctan 1 - \arctan 0} \right] \cr
& = y\left( {\frac{\pi }{4} - 0} \right) \cr
& = \frac{\pi }{4}y \cr
& \cr
& \int_0^1 {\left[ {\int_0^1 {\frac{y}{{1 + {x^2}}}} dx} \right]} dy = \int_0^1 {\frac{\pi }{4}y} dy \cr
& = \frac{\pi }{4}\int_0^1 y d \cr
& {\text{integrating}} \cr
& = \frac{\pi }{4}\left( {\frac{{{y^2}}}{2}} \right)_0^1 \cr
& = \frac{\pi }{4}\left( {\frac{{{1^2}}}{2} - \frac{{{0^2}}}{2}} \right) \cr
& {\text{simplifying}} \cr
& = \frac{\pi }{4}\left( {\frac{1}{2}} \right) \cr
& = \frac{\pi }{8} \cr} $$
