<s>[INST] 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.
 [/INST] $$\begin{align}
&\int_{-\infty}^{\infty}\frac{x^2}{(x^2+a^2)(x^2+b^2)}dx\\
=&\int^{\infty}_{0}\frac{2dx}{x^2+\frac{a^2b^2}{x^2}+(a^2+b^2)}\\
=& \int^{\infty}_{0}\frac{d(x+\frac{ab }x)}{(x+\frac{ab}{x})^2+(a-b)^2}
+ \int^{\infty}_{0}\frac{d(x-\frac{ab }x)}{(x-\frac{ab}{x})^2+(a+b)^2}\\
=& \int^{\infty}_{\infty}\frac{dt}{t^2+(a-b)^2}
+ \int^{\infty}_{-\infty}\frac{dt}{t^2+(a+b)^2}\\
=& \>0+\frac\pi{a+b}
\end{align}$$
</s>