How to compute $\int_0^\infty \frac{x^4}{(x^4+ x^2 +1)^3} dx =\frac{\pi}{48\sqrt{3}}$?

Here is an approach.

You may write

$$\begin{align} \int_0^{\infty}\frac{x^4}{\left(x^4+x^2+1\right)^3}dx &=\int_0^{\infty}\frac{x^4}{\left(x^2+\dfrac1{x^2}+1\right)^3\,x^6}dx\\\\ &=\int_0^{\infty}\frac{1}{\left(x^2+\dfrac1{x^2}+1\right)^3}\frac{dx}{x^2} \\\\ &=\int_0^{\infty}\frac{1}{\left(x^2+\dfrac1{x^2}+1\right)^3}dx\\\\ &=\frac12\int_0^{\infty}\frac{1}{\left(x^2+\dfrac1{x^2}+1\right)^3}\left(1+\dfrac1{x^2}\right)dx\\\\ &=\frac12\int_0^{\infty}\frac{1}{\left(\left(x-\dfrac1{x}\right)^2+3\right)^3}d\left(x-\dfrac1{x}\right)\\\\ &=\frac12\int_{-\infty}^{+\infty}\frac{1}{\left(u^2+3\right)^3}du\\\\ &=\frac14\:\partial_a^2\left(\left.\int_{-\infty}^{+\infty}\frac{1}{\left(u^2+a\right)}du\right)\right|_{a=3}\\\\ &=\frac14\:\partial_a^2\left.\left(\frac{\pi}{\sqrt{a}}\right)\right|_{a=3}\\\\ &=\color{blue}{\frac{\pi }{48 \sqrt{3}}} \end{align}$$

as desired.

For the antiderivative, you could also "simplify" the problem using partial fraction decomposition since $$\frac{x^4}{\left(x^4+x^2+1\right)^3}=-\frac{3 (x-1)}{16 \left(x^2-x+1\right)}+\frac{3 (x+1)}{16 \left(x^2+x+1\right)}-\frac{3}{16 \left(x^2-x+1\right)^2}-\frac{3}{16 \left(x^2+x+1\right)^2}+\frac{x}{8 \left(x^2-x+1\right)^3}-\frac{x}{8 \left(x^2+x+1\right)^3}$$ which leads to the result. But, I suspect that using residues will make the problem easier.

Could you prove that $$\int_0^1 \frac{x^4}{(x^4+ x^2 +1)^3} dx =\frac{1}{288} \left(-28+\sqrt{3} \pi +27 \log (3)\right)$$

A yet another approach:

\begin{align} I&=\int_0^\infty \frac{x^4}{(x^4+ x^2 +1)^3} dx\\ &=\int_0^1 \frac{x^4}{(x^4+ x^2 +1)^3} dx+\int_1^\infty \frac{x^4}{(x^4+ x^2 +1)^3} dx\\ &=\int_0^1\frac{x^4}{(x^4+ x^2 +1)^3} dx+\int_0^1\frac{x^6}{(x^4+ x^2 +1)^3} dx\\ \end{align} where I used $x \to 1/x$.

Now use $x\to \tan x$ to obtain \begin{align} I&=\int_0^{\pi/4} \frac{32\sin^4 2x}{(7+\cos 4x)^3} dx \end{align} which is manageable.