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?

One way:
Denote the integral as $$I=\int_{-\infty}^{\infty}\frac{ a^2x^2 dx}{(x^2-b^2)^2+a^2x^2}.$$
$$(x^2-b^2)^2+a^2x^2=(x^2+p^2)(x^2+q^2) \Rightarrow p=(a+c)/2,q=(a-c)/2, c=\sqrt{a^2-4b^2}.$$
Then $$I=\frac{2a^2}{p^2-q^2} \int_{0}^{\infty} \left( \frac{p^2}{x^2+p^2}-\frac{q^2}{x^2+q^2} \right)dx=
\frac{2a^2}{p^2-q^2}[p\tan^{-1}(x/p)-q \tan^{-1}(x/q)]_{0}^{\infty}=\frac{a^2\pi}{p+q}=a\pi.$$
