Prove that $\int_{0}^{\infty}{1\over x^4+x^2+1}dx=\int_{0}^{\infty}{1\over x^8+x^4+1}dx$ Let

$$I=\int_{0}^{\infty}{1\over x^4+x^2+1}dx\tag1$$
  $$J=\int_{0}^{\infty}{1\over x^8+x^4+1}dx\tag2$$

Prove that $I=J={\pi \over 2\sqrt3}$

Sub: $x=\tan{u}\rightarrow dx=\sec^2{u}du$
$x=\infty \rightarrow u={\pi\over 2}$, $x=0\rightarrow u=0$
Rewrite $(1)$ as
$$I=\int_{0}^{\infty}{1\over (1+x^2)^2-x^2}dx$$
then
$$\int_{0}^{\pi/2}{\sec^2{u}\over \sec^4{u}-\tan^2{u}}du\tag3$$
Simplified to
$$I=\int_{0}^{\pi/2}{1\over \sec^2{u}-\sin^2{u}}du\tag4$$
Then to
$$I=2\int_{0}^{\pi/2}{1+\cos{2u}\over (2+\sin{2u})(2-\sin{2u})}du\tag5$$
Any hints on what to do next?

Re-edit (Hint from Marco)
$${1\over x^8+x^4+1}={1\over 2}\left({x^2+1\over x^4+x^2+1}-{x^2-1\over x^4-x^2+1}\right)$$
$$M=\int_{0}^{\infty}{x^2+1\over x^4+x^2+1}dx=\int_{0}^{\infty}{x^2\over x^4+x^2+1}dx+\int_{0}^{\infty}{1\over x^4+x^2+1}dx={\pi\over \sqrt3}$$
$$N=\int_{0}^{\infty}{x^2-1\over x^4-x^2+1}dx=0$$
$$J=\int_{0}^{\infty}{1\over x^8+x^4+1}dx={1\over 2}\left({\pi\over \sqrt3}-0\right)={\pi\over 2\sqrt3}.$$
 A: Note that $$\frac{1}{x^{4}+x^{2}+1}=\frac{1}{2}\left(\frac{x+1}{x^{2}+x+1}-\frac{x-1}{x^{2}-x+1}\right)
 $$ and $$\begin{align}
\int\frac{x+1}{x^{2}+x+1}dx= & \frac{\log\left(x^{2}+x+1\right)}{2}+\frac{1}{2}\int\frac{1}{x^{2}+x+1}dx \\
 = & \frac{\log\left(x^{2}+x+1\right)}{2}+\frac{1}{2}\int\frac{1}{\left(x+\frac{1}{2}\right)^{2}+\frac{3}{4}}dx \\
 = & \frac{\log\left(x^{2}+x+1\right)}{2}+\frac{2}{3}\int\frac{1}{\left(\frac{2x+1}{\sqrt{3}}\right)^{2}+1}dx \\
 = & \frac{\log\left(x^{2}+x+1\right)}{2}+\frac{\arctan\left(\frac{2x+1}{\sqrt{3}}\right)}{\sqrt{3}}
\end{align}
 $$ and in a similar way we can compute the other integral, hence $$\begin{align}
\int_{0}^{\infty}\frac{1}{x^{4}+x^{2}+1}dx= & \frac{1}{2}\left(\log\left(\frac{\sqrt{x^{2}+x+1}}{\sqrt{x^{2}-x+1}}\right)+\frac{\arctan\left(\frac{2x+1}{\sqrt{3}}\right)+\arctan\left(\frac{2x-1}{\sqrt{3}}\right)}{\sqrt{3}}\right)_{0}^{\infty} \\ = & \frac{\pi}{2\sqrt{3}}.
\end{align}$$
For the second note that $$\frac{1}{x^{8}+x^{4}+1}=\frac{1}{4}\left(\frac{1}{x^{2}-x+1}+\frac{1}{x^{2}+x+1}\right)+\frac{1}{2}\left(\frac{x^{2}-1}{x^{4}-x^{2}+1}\right)
 $$ the first two integral are part of the previous calculation, so we have only to analyze $$I=\int_{0}^{\infty}\frac{x^{2}-1}{x^{4}-x^{2}+1}dx
 $$ and $$I=\int_{0}^{1}\frac{x^{2}-1}{x^{4}-x^{2}+1}dx+\int_{1}^{\infty}\frac{x^{2}-1}{x^{4}-x^{2}+1}dx \tag{1}
 $$ and note that if we put $x=1/y$ in the second integral of $(1)$ we get $$\int_{1}^{\infty}\frac{x^{2}-1}{x^{4}-x^{2}+1}dx=-\int_{0}^{1}\frac{y^{2}-1}{y^{4}-y^{2}+1}dy
 $$ then $$I=0.$$
A: \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 $a,c>0$
$$\int_0^\infty f\left(\left(ay-\frac{c}{y}\right)^2\right)\ \mathrm dy=\frac{1}{a}\int_0^\infty f\left(y^2\right)\ \mathrm dy$$
A: Both integrands are even, so the integrals can be extended to negative infinity. The integrands decay quickly enough for the contour to be closed with a semicircle at infinity in the upper half plane without chaning the integral. Then the integrals can be determined with the residue theorem.
Multiplying the denominator of the first integrand by $x^2-1$ yields $x^6-1$, so there are simple poles at $x=\omega_k=\exp\left(\frac{2\pi\mathrm i k}6\right)$ for $k=1,2,4,5$. The ones for $k=1,2$ lie in the upper half plane, and the corresponding residues are
$$
r_k=\lim_{x\to\omega_k}\frac{\left(x^2-1\right)\left(x-\omega_k\right)}{x^6-1}=\lim_{x\to\omega_k}\frac{(x-\omega_0)(x-\omega_3)(x-\omega_k)}{\prod_{j=0}^5(x-\omega_j)}\;,
$$
with
$$
r_1=\frac1{(\omega_1-\omega_2)(\omega_1-\omega_4)(\omega_1-\omega_5)}=-\frac14-\frac1{4\sqrt3}\mathrm i
$$
and
$$
r_2=\frac1{(\omega_2-\omega_1)(\omega_2-\omega_4)(\omega_2-\omega_5)}=\frac14-\frac1{4\sqrt3}\mathrm i
$$
Thus the contour integral is
$$
2\pi\mathrm i\left(\left(-\frac14-\frac1{4\sqrt3}\mathrm i\right)+\left(\frac14-\frac1{4\sqrt3}\mathrm i\right)\right)=\frac\pi{\sqrt3}\;,
$$
and your real integral is half of that.
For the second integral, multiplying the denominator by $x^4-1$ yields $x^{12}-1$, so there are simple poles at $\nu_k=\exp\left(\frac{2\pi\mathrm ik}{12}\right)$ for $k=1,2,4,5,7,8,10,11$, of which the ones for $k=1,2,4,5$ lie in the upper half plane. I'll leave it to you to find and add their residues as above.
A: Solution for lazy people who want to trade as much work as possible for effortless formal reasoning. Let's consider a more general integral of the form:
$$I = \int_0^{\infty}\frac{dx}{\sum_{k=0}^{N}x^{r k}}$$
Summing the geometric series in the denominator yields:
$$I = \int_0^{\infty}\frac{(1-x^r) dx}{1-x^{r(N+1)}}$$
Put $t = x^{r(N+1)}$ and define $s = \frac{1}{r(N+1)}$
$$I =s\int_0^{\infty}\frac{(1-t^{rs})t^{s-1} dt}{1-t}\tag{1}$$
Let's try to evaluate this using the standard integral:
$$\int_0^{\infty}\frac{x^{-p}dx}{1+x}=\frac{\pi}{\sin{\pi p}}\tag{2}$$
for $0<p<1$. Proving this is an easy contour integration exercise, there are lot of explanations on how to do this on this site. To evaluate (1) we need to generalize this to:
$$\int_0^{\infty}\frac{x^{-p}dx}{1+ax}=a^{p-1}\frac{\pi}{\sin{\pi p}}\tag{3}$$
Deriving (3) from (2) involves just a rescaling of the integration variable, but we'll use the r.h.s. of (3) for $a=-1$ when the integral on the l.h.s. is not defined. We'll circumvent this problem using an analytic continuation argument.
Consider generalizing the integral (1) to:
$$I(a) =s\int_0^{\infty}\frac{(1-t^{rs})t^{s-1} dt}{1+at}\tag{4}$$
Then this is well defined for $a = -1$, it is an analytic function of $a$. There is clearly no problem splitting $I(a)$ up as:
$$I(a) =s\int_0^{\infty}\frac{t^{s-1} dt}{1+at} - s\int_0^{\infty}\frac{t^{(r+1)s-1} dt}{1+at}\tag{5}$$
as long as $a\neq 1$. But note that if we substitute the expressions for these integrals using (3) then the resulting expression for $I(a)$ will be analytic in $a$, therefore by the principle of analytic continuation it will yield the correct expression also for $a = -1$. The only thing to be careful about is to use a consistent definition of the fraction power of $a$ for the two integrals. 
Using the r.h.s. of (3) for  $a = \exp(i\pi)$ yields the result:
$$I = \pi s \left[\cot(\pi s) - \cot(\pi(r+1)s)\right]$$
A: Here is an alternative approch using complex analysis. Since both integrands are even one can start with integrating around the singularities in the positive half-plane using the residue theorem which yields
$$
\begin{align*}
 \frac{1}{2}\int_{-\infty}^\infty \frac{1}{x^4+x^2+1}\,\mathrm dx
 &= \pi\mathrm i\left(\operatorname{Res}_{z=\sqrt[3]{-1}}\left(\frac{1}{z^4+z^2+1}\right) + \operatorname{Res}_{z=(-1)^{2/3}}\left(\frac{1}{z^4+z^2+1}\right)\right)\\
 &= \pi\mathrm i\left(\frac{1}{4\sqrt[3]{-1}^3+2\sqrt[3]{-1}} + \frac{1}{4((-1)^{2/3})^3+2(-1)^{2/3}}\right)\\
 &= \pi\mathrm i\left(-\frac{\mathrm i}{2\sqrt{3}}\right)\\
 &= \frac{\pi}{2\sqrt{3}}
\end{align*}
$$
and similiarly the second integral
$$
\begin{align*}
 \frac{1}{2}\int_{-\infty}^\infty \frac{1}{x^8+x^4+1}\,\mathrm dx
 &= \pi\mathrm i\left(\operatorname{Res}_{z=\sqrt[6]{-1}}\left(\frac{1}{z^8+z^4+1}\right) + \operatorname{Res}_{z=\sqrt[3]{-1}}\left(\frac{1}{z^8+z^4+1}\right)\right.\\
 &\qquad \left.+ \operatorname{Res}_{z=(-1)^{2/3}}\left(\frac{1}{z^8+z^4+1}\right) + \operatorname{Res}_{z=(-1)^{5/6}}\left(\frac{1}{z^8+z^4+1}\right)\right)\\
 &= \pi\mathrm i\left(\frac{1}{8(\sqrt[6]{-1})^7+4(\sqrt[6]{-1})^3} + \frac{1}{8(\sqrt[3]{-1})^7+4(\sqrt[3]{-1})^3}\right.\\
 &\qquad \left. + \frac{1}{8((-1)^{2/3})^7+4((-1)^{2/3})^3} +\frac{1}{8((-1)^{5/6})^7+4((-1)^{5/6})^3} \right)\\
 &= \pi\mathrm i\left(-\frac{\mathrm i}{2\sqrt{3}}\right)\\
 &= \frac{\pi}{2\sqrt{3}}
\end{align*}
$$
thus both are equal indeed.
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Leftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{I \equiv \int_{0}^{\infty}{\dd x \over x^{4} + x^{2} + 1}\,,\qquad
     J \equiv \int_{0}^{\infty}{\dd x \over x^{8} + x^{4} + 1}}$

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}
A: 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}
A: $$
\begin{aligned}
I & =\int_0^{\infty} \frac{\frac{1}{x^2}}{x^2+\frac{1}{x^2}+1} d x \\
& =\frac{1}{2} \int_0^{\infty} \frac{\left(1+\frac{1}{x^2}\right)-\left(1-\frac{1}{x^2}\right)}{x^2+\frac{1}{x^2}+1} d x \\
& =\frac{1}{2}\left[\int_0^{\infty} \frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^2+3}-\int_0^{\infty} \frac{d\left(x+\frac{1}{x}\right)}{\left(x+\frac{1}{x}\right)^2-1}\right] \\
& =\frac{1}{2}\left[\frac{1}{\sqrt{3}}\left[\tan ^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{3}}\right)\right]_0^{\infty}-\frac{1}{2}\ln \left|\frac{x+\frac{1}{x}-1}{x+\frac{1}{x}+1}\right|_0^{\infty}\right]\\
& =\frac{\pi}{2 \sqrt{3}}
\end{aligned}
$$

$$
\begin{aligned}
& J=\int_0^{\infty} \frac{1}{x^8+x^4+1} d x=\frac{1}{2}\left(\underbrace{\int_0^{\infty} \frac{x^2+1}{x^4+x^2+1}}_K d x-\underbrace{\int_0^{\infty} \frac{x^2-1}{x^4-x^2+1} d x}_L\right) \\
& K=\int_0^{\infty} \frac{1+\frac{1}{x^2}}{\left(x-\frac{1}{x}\right)^2+3} d x=\int_0^{\infty} \frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^2+1} =\frac{1}{\sqrt{3}}\left[\tan ^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{3}}\right)\right]_0^{\infty}=\frac{\pi}{\sqrt{3}} 
\end{aligned}
$$
$$
\begin{aligned}
L & =\int_0^{\infty} \frac{1-\frac{1}{x^2}}{x^2+\frac{1}{x^2}-1} d x =\int_0^{\infty} \frac{d\left(x+\frac{1}{x}\right)}{\left(x+\frac{1}{x}\right)^2-3} =\frac{1}{2 \sqrt{3}}\left[\ln \left(\frac{x+\frac{1}{x}-\sqrt{3}}{x+\frac{1}{x}+\sqrt{3}}\right)\right]_0^{\infty}  =0
\end{aligned}
$$
Hence we can conclude that$$\boxed{J=I =\frac{\pi}{2 \sqrt{3}}} $$
A: Noting that $$
J \stackrel{x\mapsto\frac{1}{x}}{=}  \int_0^{\infty} \frac{x^6}{x^8+x^4+1} d x
$$
On average,
$$
\begin{aligned}
2 J & =\int_0^{\infty} \frac{x^6+1}{x^8+x^4+1} d x \\
& =\int_0^{\infty} \frac{\left(x^2+1\right)\left(x^4-x^2+1\right)}{\left(x^4+x^2+1\right)\left(x^4-x^2+1\right)} d x \\
& =\int_0^{\infty} \frac{x^2+1}{x^4+x^2+1} d x
\end{aligned}
$$
Similarly, $$I \stackrel{x\mapsto\frac{1}{x}}{=}  \int_0^{\infty} \frac{x^2}{x^4+x^2+1} d x \Rightarrow 2I= \int_0^{\infty} \frac{x^2+1}{x^4+x^2+1} d x
$$
Hence $2I=2J $ and
$$
\begin{aligned}
I=J & =\frac{1}{2} \int_0^{\infty} \frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^2+3} \\
& =\frac{1}{\sqrt{3}}\left[\tan ^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{3}}\right)\right]_0^{\infty} \\
& =\frac{\pi}{2 \sqrt{3}}
\end{aligned}
$$
