An integral of a rational function with high degrees: Evaluate $\int_{-\infty}^{\infty}\frac{(x^4 + 1)^2}{x^{12} + 1}dx$. 
Calculate: $$\int_{-\infty}^{\infty}\frac{(x^4 + 1)^2}{x^{12} + 1}dx.$$

What I have tried is to divide both numerator and denominator with $x^4 + 1$ and then get two following integrals, because of parity of the integrand (nothing else worked for me): $$2\int_0^\infty \frac{x^4 + 1}{(x^4 - \sqrt3x^2 + 1)(x^4 + \sqrt3x^2 + 1)}dx = $$ $$ = \int_0^\infty \frac1{x^4 - \sqrt3x^2 + 1}dx + \int_0^\infty\frac1{x^4 + \sqrt3x^2 + 1}dx = $$ $$ = \int_0^\infty \frac1{(x^2 - \frac{\sqrt3}2)^2 + \frac14}dx + \int_0^\infty \frac1{(x^2 + \frac{\sqrt3}2)^2 + \frac14}dx.$$
I don't see what would be continuation of this. Any help is appreciated.
Thank you for any help. Appreciate it.

Any rational function can always be integrated by decomposing it into partial fractions. In this case this decomposition is particularly simple.
\begin{eqnarray}
\frac{1}{x^4-\sqrt{3} x^2+1} &=& \frac{1}{x^2-\theta_+} \cdot \frac{1}{\theta_+-\theta_-} + \frac{1}{x^2-\theta_-} \cdot \frac{1}{\theta_--\theta_+}\\
\frac{1}{x^4+\sqrt{3} x^2+1} &=& \frac{1}{x^2-\lambda_+} \cdot \frac{1}{\lambda_+-\lambda_-} + \frac{1}{x^2-\lambda_-} \cdot \frac{1}{\lambda_--\lambda_+} 
\end{eqnarray}
where $\theta_\pm=(\sqrt{3}\pm \imath)/2$ and $\lambda_\pm=(-\sqrt{3}\pm \imath)/2$.
Therefore 
\begin{eqnarray}
\int\limits_0^\infty \frac{dx}{x^4-\sqrt{3} x^2+1}&=& \frac{\pi}{\imath} \frac{1}{\sqrt{\frac{-\sqrt{3}-\imath}{2}}}\\
\int\limits_0^\infty \frac{dx}{x^4+\sqrt{3} x^2+1}&=& \frac{\pi}{\imath} \frac{1}{\sqrt{\frac{+\sqrt{3}-\imath}{2}}}
\end{eqnarray}
Now by using simple complex analysis we get the final result:
\begin{equation}
\int\limits_{-\infty}^\infty \frac{(x^4+1)^2}{x^{12}+1} dx = \sqrt{\frac{3}{2}}\pi
\end{equation}
