Hello everyone I hope you are doing okay.
I have been stuck in this problem that I initially thought it was an easy problem
$$\int_{(0\ , \infty)\times(0\ , 1)}{\frac{1}{1+x^2+y^2}dxdy}$$
What i have tries so far is to apply Fulbini and further work it out, after applying it i arrive at
$$\int_0^\infty\int_0^1{\frac{1}{1+x^2+y^2}dy dx}$$
Then i think of trying to take the ArcTan out with $(\frac{y}{\sqrt{1+x^2}})^2$
but then i get stuck to $$\int_0^{\infty}{\sqrt{1+x^2}}\arctan({{\sqrt{1+x^2}}})$$
and here i get stuck, I would appreciate any kind of hints.
I know the integral converges because i put it in Mathematica
Thank you in Advance
You can interchange the limits and you'll get an easier integral since $$\int_{(0\ , \infty)\times(0\ , 1)}{\frac{1}{1+x^2+y^2}dxdy}=\int_0^1\int_0^\infty{\frac{1}{1+x^2+y^2}dxdy}=\int_0^1\left(\dfrac{\arctan\left(\frac{x}{\sqrt{y^2+1}}\right)}{\sqrt{y^2+1}}\biggr|_0^\infty\right)dy=\int_0^1\frac{\pi}{2\sqrt{y^2+1}}dy=\frac{\pi}{2}\operatorname{arcsinh}y\bigg|_0^1=\frac{\pi}{2}\operatorname{arcsinh}1$$