Integrate $\int_{-\infty}^{\infty} \frac{dx}{1+x^{12}}$using partial fractions How do I integrate the improper integral  $$\int_{-\infty}^{\infty} \frac{dx}{1+x^{12}}$$ using partial fraction decomposition?
I am restricted to use only the principles taught in Calculus 2, which entails partial fraction decomposition. I have tried factoring the denominator but was quickly stumped at the complicated roots. I am unsure of how to proceed. I even thought of looking at it as a series but I don't think I can do that.
Thank you very much for your help. I greatly appreciate it.

Decompose the integrand
$$ \frac{6}{1+x^{12}}
=\frac2{x^4+1}-\frac{\sqrt3x^2-2}{x^4-\sqrt3x^2+1} 
 + \frac{\sqrt3x^2+2}{x^4+\sqrt3x^2+1} $$
Then
$$\int_{-\infty}^{\infty} \frac{dx}{1+x^{12}}
=\frac23 J(0) +\frac{2+\sqrt3}3 J(\sqrt3)+ \frac{2-\sqrt3}3 J(-\sqrt3)\tag1$$
where
\begin{align}
J(a) &= \int_{0}^{\infty} \frac{dx}{x^4+ax^2+1}
\overset{x\to \frac1x} =\int_{0}^{\infty} \frac{x^2dx}{x^4+ax^2+1} \\
& =\frac12 \int_{0}^{\infty} \frac{(x^2+1)dx}{x^4+ax^2+1}
  =\frac12 \int_{0}^{\infty} \frac{d(x-\frac1x)}{(x-\frac1x)^2+2+a} 
=\frac\pi{2\sqrt{2+a}}\\
\end{align}
Let $a=0, \pm \sqrt3$ and plug the results into (1) to obtain
$$\int_{-\infty}^{\infty} \frac{dx}{1+x^{12}} = \frac{\sqrt3+1}{3\sqrt2}$$
