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?
Answer
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.
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