Need help with $\int_{-\infty}^\infty \frac{x^2 \, dx}{x^4+2a^2x^2+b^4}$ I'm having trouble trying to evaluate this definite integral. Mathematica didn't help much.
$$\int_{-\infty}^\infty \frac{x^2 \, dx}{x^4+2a^2x^2+b^4}$$
where $a$, $b$ $\in \Bbb R^+$.
Is it possible to solve it analytically? What methods should I try?

We can write the integrand as
$$\begin{align}
\frac{x^2}{x^4+2a^2x^2+b^4}&=\frac{x^2}{(x^2+a^2+\sqrt{a^4-b^4})(x^2+a^2-\sqrt{a^4-b^4})}\\\\
&=\frac{A}{x^2+a^2-\sqrt{a^4-b^4}}+\frac{B}{x^2+a^2+\sqrt{a^4-b^4}}
\end{align}$$
where $A$ and $B$ are given respectively by
$$A=\frac{-a^2+\sqrt{a^4-b^4}}{2\sqrt{a^4-b^4}}$$
and
$$B=\frac{a^2+\sqrt{a^4-b^4}}{2\sqrt{a^4-b^4}}$$
Therefore, we find that for $a>b>0$
$$\begin{align}
\int_{-\infty}^\infty \frac{x^2}{x^4+2a^2x^2+b^4}\,dx&=2\int_0^\infty \frac{x^2}{x^4+2a^2x^2+b^4}\,dx\\\\
&=2A\int_0^\infty \frac{1}{x^2+a^2-\sqrt{a^4-b^4}}\,dx+2B\int_0^\infty \frac{1}{x^2+a^2+\sqrt{a^4-b^4}}\,dx\\\\
&=\frac{\pi A}{\sqrt{a^2-\sqrt{a^4-b^4}}}+\frac{\pi B}{\sqrt{a^2+\sqrt{a^4-b^4}}}\\\\
&=\frac{\pi}{2\sqrt{a^4-b^4}}\left(\sqrt{a^2+\sqrt{a^4-b^4}}-\sqrt{a^2-\sqrt{a^4-b^4}}\right)\\\\
&=\frac{\pi}{2\sqrt{a^4-b^4}}\left(\sqrt{2(a^2-b^2)}\right)\\\\
&=\frac{\pi}{\sqrt{2(a^2+b^2)}}
\end{align}$$
as expected!

Note that for $b>a>0$, result of the preceding development is unaltered.  One needs only to proceed with caution when evaluating the integrals, $\int \frac{1}{x^2+a^2\pm \sqrt{a^4-b^4}}\,dx$ due to the complex constants $a^2\pm \sqrt{a^4-b^4}$.
For the case $a=b>0$, the result of the preceding developing is unaltered.  The integral $\int_0^\infty \frac{x^2}{(x^2+a^2)^2}\,dx$ can easily be evaluated enforcing the substitution $x\to a\tan(x)$. 

