A tricky double integral What is
$$\int_0^1 \int_0^1 \frac{ dx \; dy}{1+xy+x^2y^2} ? $$
Can you do one of the integrals and turn it into a single integral?
I get lost in a sea of inverse tangents.

The first step would be one $\arctan$ only. 
The denominator is 
$$1+xy+x^2y^2 =\left(\frac{1}{2}+xy\right)^2 + \frac{3}{4}$$
Hence, with $a=\frac{2}{\sqrt{3}}$ we have
$$\int\frac{dy}{1+xy+x^2y^2} =\int\frac{dy}{\left(\frac{1}{2}+xy\right)^2 + 1/a^2}=\qquad\qquad\qquad\text{               }\\
a^2\int\frac{dy}{a^2\left(\frac{1}{2}+xy\right)^2 + 1}=
a^2\frac{\arctan(\frac{a}{2}+axy)}{ax} + C=\\a\frac{\arctan\left(\frac{1}{\sqrt{3}}+\frac{2xy}{\sqrt{3}}\right)}{x} + C$$
so that $$\int_0^1\frac{dy}{1+xy+x^2y^2} = a\frac{\arctan(\frac{1}{\sqrt{3}}+\frac{2x}{\sqrt{3}})}{x} -\frac{\pi}{3\sqrt{3}}\frac{1}{x}$$
The integral of $\arctan(a+bx)/x$ is not that easy, it ends up in dilogarithms... 
