Prove that $\int_{0}^{\infty}{1\over x^4+x^2+1}dx=\int_{0}^{\infty}{1\over x^8+x^4+1}dx$
$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{I \equiv \int_{0}^{\infty}{\dd x \over x^{4} + x^{2} + 1}\,,\qquad J \equiv \int_{0}^{\infty}{\dd x \over x^{8} + x^{4} + 1}}$
Note that $\ds{x^{8} + x^{4} + 1 = \pars{x^{4} + x^{2} + 1}\pars{x^{4} - x^{2} + 1}}$ such that \begin{align} I - J & = \int_{0}^{\infty}{x^{4} - x^{2} \over x^{8} + x^{4} + 1}\,\dd x\ \stackrel{x\ \to\ 1/x}{=}\ \int_{\infty}^{0}{1/x^{4} - 1/x^{2} \over 1/x^{8} + 1/x^{4} + 1} \,{\dd x \over -x^{2}} = \int_{0}^{\infty}{x^{2} - x^{4} \over x^{8} + x^{4} + 1}\,\dd x \\[3mm] & = J - I\quad\imp\quad \fbox{$\ds{\quad\color{#f00}{I} = \color{#f00}{J}\quad}$} \end{align}
The problem is reduced to evaluate $\ds{\underline{just\ one}}$ of the above integrals: For example, $\ds{\color{#f00}{I}}$. \begin{align} \color{#f00}{I} & = \int_{0}^{\infty}{\dd x \over x^{4} + x^{2} + 1}\ \stackrel{x\ \to\ 1/x}{=}\ \int_{0}^{\infty}{\dd x \over 1/x^{2} + 1 + x^{2}} = \int_{0}^{\infty}{\dd x \over \pars{x - 1/x}^{2} + 3}\tag{1} \\[3mm] & \mbox{Similarly,} \\[3mm] \color{#f00}{I} & = \int_{0}^{\infty}{\dd x \over x^{4} + x^{2} + 1} = \int_{0}^{\infty}{1 \over x^{2} + 1 + 1/x^{2}}\,{\dd x \over x^{2}} = \int_{0}^{\infty}{1 \over \pars{x - 1/x}^{2} + 3} \,\dd\pars{-\,{ 1\over x}}\tag{2} \end{align}
With $\pars{1}$ and $\pars{2}$: \begin{align} \color{#f00}{I} & = \color{#f00}{J} = \half\int_{x = 0}^{x \to \infty}{1 \over \pars{x - 1/x}^{2} + 3} \,\dd\pars{x - {1 \over x}}\ \stackrel{\pars{x - 1/x}\ \to x}{=}\ \half\int_{-\infty}^{\infty}{\dd x \over x^{2} + 3} \\[3mm] \stackrel{x/\root{3}\ \to\ x}{=}\ &\ {1 \over \root{3}}\ \underbrace{\int_{0}^{\infty}{\dd x \over x^{2} + 1}}_{\ds{=\ {\pi \over 2}}}\ =\ \color{#f00}{\pi \over 2\root{3}} \end{align}
For $\theta$ an arbitrary constant, we have \begin{equation} x^4+2x^2\cos2\theta+1=(x^2-2x\sin\theta+1)(x^2+2x\sin\theta+1)\tag1 \end{equation} Now, let’s evaluate \begin{equation} I=\int_0^\infty\frac{1}{x^4+2x^2\cos2\theta+1}\ dx\tag2 \end{equation} By making substitution $x\mapsto\frac1x$, then \begin{equation} I=\int_{0}^\infty\frac{x^2}{x^4+2x^2\cos2\theta+1}\ dx\tag3 \end{equation} Adding $(2)$ and $(3)$, we get \begin{equation} I=\frac12\int_{0}^\infty\frac{1+x^2}{x^4+2x^2\cos2\theta+1}\ dx=\frac14\int_{-\infty}^\infty\frac{1+x^2}{x^4+2x^2\cos2\theta+1}\ dx\tag4 \end{equation} Since the integrand is even, an odd function like $2x\sin\theta$ doesn't change the value of the integral, so by using $(1)$ we have \begin{align} I&=\frac14\int_{-\infty}^\infty\frac{x^2+2x\sin\theta+1}{x^4+2x^2\cos2\theta+1}\ dx\\[10pt] &=\frac14\int_{-\infty}^\infty\frac{1}{x^2-2x\sin\theta+1}\ dx\\[10pt] &=\frac14\int_{-\infty}^\infty\frac{1}{(x-\sin\theta)^2+\cos^2\theta}\ dx\\[10pt] &=\frac1{4\cos\theta}\int_{-\infty}^\infty\frac{1}{y^2+1}\ dy\quad\longrightarrow\quad y={x-\sin\theta\over\cos\theta}\\[10pt] &=\frac\pi{4\cos\theta} \end{align} For $\theta=\frac\pi3$, then \begin{equation} \int_0^\infty\frac{1}{x^4+x^2+1}\ dx=\frac{\pi}{2\sqrt3} \end{equation} as announced.
For $J$ we use \begin{equation} x^8+2x^4\cos2\theta+1=(x^4-2x^2\sin\theta+1)(x^4-2x^2\sin\theta+1) \end{equation} Then apply the same methods for \begin{equation} J=\int_0^\infty\frac{1}{x^8+2x^4\cos2\theta+1}\ dx \end{equation}
\begin{align} \int_0^{\infty}\frac{\mathrm dx}{a^2x^4+bx^2+c^2} &= \int_0^{\infty}\frac{\mathrm dx}{\left(ax-\frac{c}{x}\right)^2+b+2ac}\cdot \frac{1}{x^2}\\[9pt] &=\frac{c}{a} \underbrace{\int_0^{\infty}\frac{\mathrm dy}{\left(ay-\frac{c}{y}\right)^2+b+2ac}}_{\large\color{blue}{x=\frac{c}{ay}}}\\[9pt] &=\frac{c}{a} \underbrace{\int_0^{\infty}\frac{\mathrm dy}{y^2+b+2ac}}_{\large\color{blue}{y=z\sqrt{b+2ac}}}\tag{$\spadesuit$}\\[9pt] &=\frac{c}{a\sqrt{b+2ac}} \int_0^{\infty}\frac{\mathrm dz}{z^2+1}\\[9pt] &=\bbox[8pt,border:3px #FF69B4 solid]{\color{red}{\large\frac{c\pi}{2a\sqrt{b+2ac}}}} \end{align}
Setting $a=b=c=1$, then
$$I=\int_0^{\infty}\frac{\mathrm dx}{x^4+x^2+1}=\bbox[8pt,border:3px #FF69B4 solid]{\color{red}{\large\frac{\pi}{2\sqrt{3}}}}$$
$(\spadesuit)$ Cauchy-Schlomilch transformation for $f(\cdot)$ is a continuous function and $a,c>0$
$$\int_0^\infty f\left(\left(ay-\frac{c}{y}\right)^2\right)\ \mathrm dy=\frac{1}{a}\int_0^\infty f\left(y^2\right)\ \mathrm dy$$