How to compute $\int_0^\infty \frac{x^4}{(x^4+ x^2 +1)^3} dx =\frac{\pi}{48\sqrt{3}}$? $$\int_0^\infty \frac{x^4}{(x^4+ x^2 +1)^3} dx =\frac{\pi}{48\sqrt{3}}$$
I have difficulty to evaluating above integrals. 
First I try the substitution $x^4 =t$ or $x^4 +x^2+1 =t$ but it makes integral worse. 
Using Mathematica I found the result $\dfrac{\pi}{48\sqrt{3}}$ I want to know the procedure of evaluating this integral. 
 A: 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: 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.
A: 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.
