<s>[INST] 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.
 [/INST] 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}
</s>