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Ho to find the following integral $$\int_{-\infty}^{\infty}\frac{1}{x^{12} + 1} dx $$using parts by substitution, partial fractions, etc. but not Cauchy's residue theorem?

Venus
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Mambo
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  • http://math.stackexchange.com/questions/110494/possibility-to-simplify-sum-limits-k-infty-infty-frac-left http://math.stackexchange.com/questions/110457/closed-form-for-int-0-infty-fracxn1-xmdx – Pedro Dec 06 '14 at 19:58
  • without gamma functions using school level integration? techniques – Mambo Dec 06 '14 at 20:02
  • Use the symmetry of the integrand, let $t=\dfrac1{x^{12}+1}$, and recognize the expression of the beta function in the new integral, then employ Euler's reflection formula for the $\Gamma$ function. – Lucian Dec 06 '14 at 20:07
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    @Lucian elementarily? – Mambo Dec 06 '14 at 20:09
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    Factor the denominator as follows: $$x^2+1~=~\Big(x^4\Big)^3+1^3~=~\ldots$$ and use partial fractions. Then factor the two new denominators as well: $$x^4+1~=~\Big(x^2+1\Big)^2-2x^2~=~\Big(x^2-x\sqrt2+1\Big) \Big(x^2+x\sqrt2+1\Big)$$ and $$x^8-x^4+1~=~\Big(x^4+1\Big)^2-3x^4~=~\Big(x^4-x^2\sqrt3+1\Big) \Big(x^4+x^2\sqrt3+1\Big)$$ – Lucian Dec 06 '14 at 20:23

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Following Lucian's comment, by partial fractions we have $$\frac{1}{x^{12}+1}=\frac{1}{\left(x^4+1\right) \left(x^8-x^4+1\right)}=\frac{1}{3}\left[\frac{1}{x^4+1}+\frac{2}{x^8-x^4+1}-\frac{x^4}{x^8-x^4+1}\right]$$ where

$$\begin{align}\frac{1}{x^4+1}=&\,\frac{x-\sqrt{2}}{2 \sqrt{2} \left(-x^2+\sqrt{2} x-1\right)}+\frac{x+\sqrt{2}}{2 \sqrt{2} \left(x^2+\sqrt{2} x+1\right)}\\[15pt]\frac{1}{x^8-x^4+1}=&\,\frac{2 x-\sqrt{6}}{2 \sqrt{2} \left(2+\sqrt{3}\right) \left(2 \sqrt{3}-3\right) \left(-2 x^2+\sqrt{2} x+\sqrt{6} x-2\right)}\\&\,+\frac{2 x-\sqrt{6}}{2 \sqrt{2} \left(2+\sqrt{3}\right) \left(2 \sqrt{3}-3\right) \left(-2 x^2-\sqrt{2} x+\sqrt{6} x-2\right)}\\&\,+\frac{2 x+\sqrt{6}}{2 \sqrt{2} \left(2+\sqrt{3}\right) \left(2 \sqrt{3}-3\right) \left(2 x^2+\sqrt{2} x+\sqrt{6} x+2\right)}\\&\,+\frac{2 x+\sqrt{6}}{2 \sqrt{2} \left(2+\sqrt{3}\right) \left(2 \sqrt{3}-3\right) \left(2 x^2-\sqrt{2} x+\sqrt{6} x+2\right)}\\[15pt]\frac{x^4}{x^8-x^4+1}=&\,\frac{\left(1+\sqrt{3}\right) x}{2 \sqrt{2} \left(2+\sqrt{3}\right) \left(2 \sqrt{3}-3\right) \left(-2 x^2-\sqrt{2} x+\sqrt{6} x-2\right)}\\&\,+\frac{\left(1+\sqrt{3}\right) x}{2 \sqrt{2} \left(2+\sqrt{3}\right) \left(2 \sqrt{3}-3\right) \left(2 x^2-\sqrt{2} x+\sqrt{6} x+2\right)}\\&\,+\frac{\left(\sqrt{3}-3\right) x}{2 \sqrt{6} \left(2+\sqrt{3}\right) \left(2 \sqrt{3}-3\right) \left(-2 x^2+\sqrt{2} x+\sqrt{6} x-2\right)}\\&\,-\frac{\left(\sqrt{3}-1\right) x}{2 \sqrt{2} \left(2+\sqrt{3}\right) \left(2 \sqrt{3}-3\right) \left(2 x^2+\sqrt{2} x+\sqrt{6} x+2\right)}\end{align}$$ The rest evaluations are elementary but tedious.

Venus
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