Integrate $int_{-infty}^{infty} frac{dx}{1+x^{12}}$using partial fractions

Question

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.

Quanto · Answer

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{xto 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} 
=fracpi{2sqrt{2+a}}\
end{align}
Let $a=0, pm sqrt3$ and plug the results into (1) to obtain
$$int_{-infty}^{infty} frac{dx}{1+x^{12}} = frac{sqrt3+1}{3sqrt2}$$

Frank Newman · Answer

To help you get started:
$u^4 + 1 =
u^4 + 2u^2 + 1 - 2u^2 $. 
Then write as a difference of squares and factor.

Integrate $int_{-infty}^{infty} frac{dx}{1+x^{12}}$using partial fractions

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