Answer
$\int_0^1 \int_{-\sqrt {1-y^2}}^{\sqrt {1-y^2}} \dfrac{y}{(1+x^2+y^2)^2} \ dx \ dy=\dfrac{\pi}{4}-\dfrac{1}{2}$
Work Step by Step
Here, we have: $\int_0^1 \int_{-\sqrt {1-y^2}}^{\sqrt {1-y^2}} \dfrac{y}{(1+x^2+y^2)^2} \ dx \ dy=\int_0^1 \int_{-\sqrt {1-y^2}}^{\sqrt {1-y^2}} \dfrac{y}{(1+x^2+y^2)^2} \ dy \ dx$
or, $=\int_{-1}^1 [-\dfrac{1}{2} \times (1+x^2+y^2)^{-1}]_0^{\sqrt {1-x^2}} \ dx $
or, $=\int_{-1}^1 [\dfrac{-1}{4}+\dfrac{1}{2(1+x^2)}] \ dx$
or, $=[\dfrac{-1}{4}x+\dfrac{1}{2} \tan^{-1} x]_{-1}^1$
or, $=\dfrac{-1}{4}x+\dfrac{1}{2}\times \dfrac{\pi}{4}-\dfrac{1}{4}+\dfrac{1}{2} \times \dfrac{\pi}{4}$
Thus, we get:
$\int_0^1 \int_{-\sqrt {1-y^2}}^{\sqrt {1-y^2}} \dfrac{y}{(1+x^2+y^2)^2} \ dx \ dy=\dfrac{\pi}{4}-\dfrac{1}{2}$
