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.

Note that we can write 
$$\begin{align}
\int_{-\infty}^\infty \frac{x^2}{(x^2+a^2)(x^2+b^2)}\,dx&=\frac12\int_{-\infty}^\infty \frac{(x^2+b^2)+(x^2+a^2)}{(x^2+a^2)(x^2+b^2)}\,dx-\frac{1}2\int_{-\infty}^\infty\frac{a^2+b^2}{(x^2+a^2)(x^2+b^2)}\,dx\\\\
&=\frac12\int_{-\infty}^\infty \frac1{x^2+a^2}\,dx+\frac12\int_{-\infty}^\infty \frac1{x^2+b^2}\,dx\\\\
&-\frac{a^2+b^2}2\int_{-\infty}^\infty\frac{1}{(x^2+a^2)(x^2+b^2)}\,dx\\\\
&=\frac\pi {2a}+\frac\pi {2b} -\frac{a^2+b^2}2\int_{-\infty}^\infty\frac{1}{(x^2+a^2)(x^2+b^2)}\,dx\tag1
\end{align}$$
Now apply Parseval to the integral on the right-hand side of $(1)$ with $f(x)=\frac{1}{x^2+a^2}$ and $g(x)=\frac1{x^2+b^2}$ and $F(k)=\frac{\pi}{|a|}e^{-|a|k}$ and $G(k)=\frac\pi{|b|}e^{-|b|k}$.
