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. 

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