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 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.
