$\int \frac{x^4-1}{x^2\sqrt{x^4+x^2+1}}dx$ $$\int \frac{x^4-1}{x^2\sqrt{x^4+x^2+1}}\,dx$$
My attempt, 
I changed it to $\int \frac{x^2}{\sqrt{x^4+x^2+1}} \, dx-\int \frac 1 {x^2\sqrt{x^4+x^2+1}} \, dx$, but I stuck here.
Any method to solve this integral? Thanks in advance.

For $x>0,$
$$I=\int \frac{x^4-1}{x^2\sqrt{x^4+x^2+1}}\,dx=\int\dfrac{x^3\left(2x-\dfrac2{x^3}\right)\ dx}{2x^3\sqrt{x^2+\dfrac1{x^2}+1}}$$
Set $x^2+\dfrac1{x^2}+1=y$
