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?
 A: 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 $$I=\frac{2a^2}{p^2-q^2} \int_{0}^{\infty} \left( \frac{p^2}{x^2+p^2}-\frac{q^2}{x^2+q^2} \right)dx=
\frac{2a^2}{p^2-q^2}[p\tan^{-1}(x/p)-q \tan^{-1}(x/q)]_{0}^{\infty}=\frac{a^2\pi}{p+q}=a\pi.$$
A: Glasser's master theorem states that for arbitrary constants $\alpha$, $(\alpha_n)_{n=1}^{N}$, $(\beta_n)_{n=1}^N$, the function 
$$\phi(x)=|\alpha|x-\sum_{n=1}^{N}\frac{|\alpha_n|}{x-\beta_n},$$
and any integrable function $F(x)$, 
$$\mathrm{PV}\int_{-\infty}^{\infty}F(\phi(x))dx=\mathrm{PV}\int_{-\infty}^\infty F(x)dx.$$
For your integral, set $$\phi(x)=x-\frac{b^2}{x}$$
and 
$$F(\phi(x))=\frac{a^2x^2}{(x^2-b^2)^2-a^2x^2}=\frac{1}{(\frac{x}{a}-\frac{b^2}{ax})^2+1}$$
to immediately yield the desired result.
A: 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
$$I=\frac{a^2}{r^2-s^2} 2i\pi \left( r^2  Res \left (\frac{1}{x^2-r^2} \right)_{x=r}-s^2  Res \left ( \frac{1}{x^2-s^2} \right)_{x=-s}\right)=\frac{i\pi a^2}{r-s}=a\pi.$$
