Whats the result of double integral $\int_{0}^{1}\int_{0}^{1}\frac{1+x^2}{1+x^2+y^2}dxdy$ Whats the result of double integral
 $$\int_{0}^{1}\int_{0}^{1}\frac{1+x^2}{1+x^2+y^2}dxdy$$

I was trying to get this $$\int_{0}^{1}\sqrt{1+x^2}\arctan {\frac{1}{\sqrt{1+x^2}}}dx=\int_{0}^{1}(1+x^2)\frac{1}{\sqrt{1+x^2}}\arctan {\frac{1}{\sqrt{1+x^2}}}dx$$
$$=\int_{0}^{1}(1+x^2)\int_{0}^{1}\frac{1}{1+x^2+y^2}dxdy=\int_{0}^{1}\int_{0}^{1}\frac{1+x^2}{1+x^2+y^2}dxdy$$
So far I don't know what to do next.any helps are to be grateful.
By symmetry one has
$$Q:=\int_0^1\int_0^1{1+x^2\over1+x^2+y^2}\&gt;dy\&gt;dx={1\over2}\int_0^1\int_0^1{2+x^2+y^2\over1+x^2+y^2}\&gt;dy\&gt;dx$$
and therefore
$$Q={1\over2}\int_0^1\int_0^1\left(1+{1\over 1+x^2+y^2}\right)\&gt;dy\&gt;dx={1\over2}+\int_0^{\pi/4}\int_0^{\sec\phi}{r\over 1+r^2}\&gt;dr\&gt;d\phi\ .$$
The last inner integral evaluates to $${1\over2}\log(1+r^2)\biggr|_0^{\sec\phi}={1\over2}\log(1+\sec^2\phi)\ ,$$
so that we finally obtain
$$Q={1\over2}+{1\over2}\int_0^{\pi/4}\log(1+\sec^2\phi)\&gt;d\phi\doteq0.819755\ .$$
(Mathematica can evaluate the last integral in terms of arcsines and polylogs.)
Changing the variable $u=1+x^2$, integrating for y, and reversing the change, you can get:
$$I = \int(\sqrt{(x^2+1)}\arctan(\frac 1 {\sqrt{(x^2+1)}}), x = 0..1 ) = 0.819755$$