Showing that $\int_{-\infty}^{\infty}\frac{x^2}{(x^2+a^2)(x^2+b^2)}dx=\frac{\pi}{a+b}$ via Fourier Transform If $a,b>0$, how can I prove this using Fourier Series
$$\int_{-\infty}^{\infty}\frac{x^2}{(x^2+a^2)(x^2+b^2)}dx=\frac{\pi}{a+b}.$$
I tried to split the product and calculate the integral using Parceval's Theorem, but $\frac{x}{(x^2+a^2)}$ and $\frac{x^2}{(x^2+a^2)}$ aren't in $L^1(\mathbb{R})$.
Any hints hold be appreciated.

HINT: Unless you’re dead-set on using the Fourier transform, I would try using that
$$\frac{x^2}{(x^2+a^2)(x^2+b^2)}=\frac{1}{a^2-b^2}\bigg(\frac{a^2}{x^2+a^2}-\frac{b^2}{x^2+b^2}\bigg)$$
