<s>[INST] The integral $\int_{-\infty}^{\infty}\frac{ a^2x^2 dx}{(x^2-b^2)^2+a^2x^2}=a\pi,~ a, b \in \Re ?$ This integral 
$$\int_{-\infty}^{\infty}\frac{ a^2x^2 dx}{(x^2-b^2)^2+a^2x^2}=a\pi, ~ a, b \in \Re$$
looks suspiciously interesting as it is independent of the parameter $b$. The question is: What is the best way of proving or disproving this?
 [/INST] How about this, please check it critically
Denote the integral as $$I=\int_{-\infty}^{\infty}\frac{ a^2x^2 dx}{(x^2-b^2)^2+a^2x^2}.$$
$$(x^2-b^2)^2+a^2x^2=(x^2-r^2)(x^2-s^2) \Rightarrow r=(d+ia)/2,s=(d-ia)/2, d=\sqrt{4b^2-a^2}.$$
$$I=\frac{a^2}{r^2-s^2} \int_{-\infty}^{\infty} \left( \frac{r^2}{x^2-r^2}-\frac{s^2}{x^2-s^2} \right)dx.$$
By considering a semi-circle contour in the upper half plane and applying the residue theorem, we get
$$I=\frac{a^2}{r^2-s^2} 2i\pi \left( r^2  Res \left (\frac{1}{x^2-r^2} \right)_{x=r}-s^2  Res \left ( \frac{1}{x^2-s^2} \right)_{x=-s}\right)=\frac{i\pi a^2}{r-s}=a\pi.$$
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