# How to 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.

• This doesn't have any roots over $\Bbb R$ so a partial fraction decomposition is going to be pretty challenging. Wolfram Alpha gives the solution in terms of a product of gamma functions as well. – Cameron Williams May 16 '15 at 1:41
• Using half-angle identities one can compute the complex roots $e^{ik\pi/12} = \cos(k\pi/12) + i\sin(k\pi/12)$ to get the roots over $\Bbb C$, then combine the conjugate terms to get the factorization of the denominator over $\Bbb R$. This is ugly and I don't see any nicer way off the top of my head. – Rolf Hoyer May 16 '15 at 1:45
• Parentheses, please. What you wrote is the integral of $1+x^{12}$, but I am sure you meant $1/(1+x^{12})$ If you know where the roots of $-1$ are, you can group them in conjugate pairs to get a factorization of $1+x^{12}$ into real quadratics – Ross Millikan May 16 '15 at 1:45
• the factorization is $(x^2\pm\sqrt{2}x+1)(x^2\pm\frac{\sqrt{3}-1}{\sqrt{2}}x+1)(x^2\pm\frac{\sqrt{3}+1}{\sqrt{2}}x+1)$ ($\pm$ are for short, every parenthesis counts 2 times, for $+$ and for $-$) @Sam – Alexey Burdin May 16 '15 at 1:49
• Someone edit this question – Ilaya Raja S May 16 '15 at 1:57

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