Double Delight

Changing the order of integration:

$\int_0^\infty \int_1^\infty \frac{x}{(1+x^2)(1+x^2y^2)} \mathrm{d}x\mathrm{d}y$

$=\int_0^\infty \frac{1}{x(1+x^2)} \int_1^\infty\frac1{(y^2+x^{-2})} \mathrm{d}y\mathrm{d}x$

$=\int_0^\infty \frac{1}{x(1+x^2)} [x\arctan(xy)]_{y=1}^{y\rightarrow\infty}\mathrm{d}x$

$=\int_0^\infty \frac{1}{1+x^2} (\frac{\pi}2-\arctan x)\mathrm{d}x$

$=[\frac{\pi}2\arctan x - \frac12 (\arctan x)^2] _{x=0}^{x\rightarrow\infty}$

$=\left(\frac{\pi}2\right)^2-\frac12\left(\frac{\pi}2\right)^2$

$=\frac{\pi^2}8$

Hence $a+b=2+8=\boxed{10}$ .