<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] 
Hint: For integrals of rational functions of even polynomials over the real line, the following substitution can come in handy:
$$\begin{align}
\mathcal{I}
&=\int_{-\infty}^{\infty}\mathrm{d}x\,\frac{\left(x^{4}+1\right)^{2}}{x^{12}+1}\\
&=\int_{-\infty}^{\infty}\mathrm{d}x\,\frac{\left(x^{4}+1\right)^{2}}{\left(x^{4}+1\right)\left(x^{8}-x^{4}+1\right)}\\
&=\int_{-\infty}^{\infty}\mathrm{d}x\,\frac{x^{4}+1}{x^{8}-x^{4}+1}\\
&=2\int_{0}^{\infty}\mathrm{d}x\,\frac{x^{4}+1}{x^{8}-x^{4}+1}\\
&=2\int_{0}^{\infty}\mathrm{d}x\,\frac{\left(x^{2}+x^{-2}\right)}{x^{2}\left(x^{4}+x^{-4}-1\right)}\\
&=2\int_{0}^{\infty}\mathrm{d}x\,\frac{\left(x^{2}+x^{-2}\right)}{\left(x^{4}+x^{-4}-1\right)};~~~\small{\left[x\mapsto\frac{1}{x}\right]}\\
&=\int_{0}^{\infty}\mathrm{d}x\,\frac{\left(x^{2}+x^{-2}\right)}{x^{2}\left(x^{4}+x^{-4}-1\right)}+\int_{0}^{\infty}\mathrm{d}x\,\frac{\left(x^{2}+x^{-2}\right)}{\left(x^{4}+x^{-4}-1\right)}\\
&=\int_{0}^{\infty}\mathrm{d}x\,\frac{x^{2}+1}{x^{2}}\cdot\frac{\left(x^{2}+x^{-2}\right)}{\left(x^{4}+x^{-4}-1\right)}\\
&=\int_{-\infty}^{\infty}\mathrm{d}y\,\frac{2\left(4y^{2}+2\right)}{\left(16y^{4}+16y^{2}+1\right)};~~~\small{\left[\frac{x^{-1}-x}{2}=y\iff x=\sqrt{y^{2}+1}-y\right]}.\\
\end{align}$$
Note that we have cut the degrees of the polynomials in half and still have all integer coefficients. This could be considered an advantage over the brute-force partial fraction method if you're looking to avoid messy factorizations.
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