TITLE: Prove that $\int_{-\infty}^{+\infty} \frac{1}{x^4+x^2+1}dx = \frac{\pi}{\sqrt{3}}$
QUESTION [7 upvotes]: Good evening everyone,
how can I prove that
$$\int_{-\infty}^{+\infty} \frac{1}{x^4+x^2+1}dx = \frac{\pi}{\sqrt{3}}\;?$$
Well, I know that $\displaystyle\frac{1}{x^4+x^2+1} $ is an even function and the interval $(-\infty,+\infty)$ is symmetric about zero, so
$$\int_{-\infty}^{+\infty} \frac{1}{x^4+x^2+1}dx = 2\int_0^\infty \frac{1}{x^4+x^2+1}dx.$$
Then I use the partial fraction:
$$2\int_0^\infty \frac{1}{x^4+x^2+1}dx= 2\int_0^\infty \left( \frac{1-x}{2(x^2-x+1)} + \frac{x+1}{2(x^2+x+1)} \right)dx.$$
So that's all. What's next step?

REPLY [3 votes]: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{\int_{-\infty}^{\infty}{1 \over x^{4} + x^{2} + 1}\,\dd x
     ={\pi \over \root{3}}\;?}$

\begin{align}&\color{#66f}{\large%
\int_{-\infty}^{\infty}{1 \over x^{4} + x^{2} + 1}\,\dd x}
=2\int_{0}^{\infty}\ {1 \over x^{2} + 1 + 1/x^{2}}\,{1 \over x^{2}}\,\dd x\
=\ \overbrace{%
2\int_{0}^{\infty}\ {1 \over \pars{x - 1/x}^{2} + 3}\,{1 \over x^{2}}\,\dd x}
^{\dsc{I_{1}}}
\\[5mm]&\stackrel{\dsc{x}\ \mapsto\ \dsc{1 \over x}}{=}\ \overbrace{%
2\int_{0}^{\infty}\ {1 \over \pars{x - 1/x}^{2} + 3}\,\dd x}
^{\dsc{I_{2}}}
\end{align}

Then,
\begin{align}&\color{#66f}{\large%
\int_{-\infty}^{\infty}{1 \over x^{4} + x^{2} + 1}\,\dd x}
={\dsc{I_{1}} + \dsc{I_{2}} \over 2}\ =\ \overbrace{%
\int_{0}^{\infty}\ {1 \over \pars{x - 1/x}^{2} + 3}\pars{{1 \over x^{2}} + 1}
\,\dd x}^{\ds{\dsc{x - {1 \over x}}\ \mapsto\ \dsc{x}}}
\\[5mm]&=\int_{-\infty}^{\infty}\ {1 \over x^{2} + 3}\,\dd x
={1 \over \root{3}}\
\overbrace{\int_{-\infty}^{\infty}\ {1 \over x^{2} + 1}\,\dd x}^{\ds{=\ \dsc{\pi}}}
\ =\ \color{#66f}{\large{\pi \over \root{3}}}
\end{align}