$\int_0^\infty {dx\over (x^2+a^2)(x^2+b^2)}={\pi\over2ab(a+b)}$ Show that $$\int_0^\infty {dx\over (x^2+a^2)(x^2+b^2)}={\pi\over2ab(a+b)}$$ where $a,b>0$. I'm not sure how to simplify this. Any solutions or hints are greatly appreciated.

You should take the partial fraction of $\dfrac{1}{(x^2+a^2)(x^2+b^2)}$.
This way, it is easier for you to integrate the expression and you should expect the $\arctan$ function in the numerator.
Recall:
$\dfrac{1}{(x^2+a^2)(x^2+b^2)}=\dfrac{Ax+B}{x^2+a^2}+\dfrac{Cx+D}{x^2+b^2}$.
