$\int \frac{x^4-1}{x^2\sqrt{x^4+x^2+1}}dx$ $$\int \frac{x^4-1}{x^2\sqrt{x^4+x^2+1}}\,dx$$
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.

Everytime I see an integrand that look sort of "symmetric" under $x \to x^{-1}$,
I will try to see whether it can be reexpressed in terms of $x \pm x^{-1}$.
One useful identity to remember is 
$$\frac{dx}{x} = \frac{d(x+x^{-1})}{x - x^{-1}} = \frac{d(x-x^{-1})}{x + x^{-1}}$$
This trick does work for the integral at hand.
$$\begin{align}
\int \frac{x^4-1}{x^2\sqrt{x^4+x^2+1}} dx &= \int \frac{x^2-x^{-2}}{\sqrt{x^2+x^{-2}+1}} \frac{dx}{x} \\
&= \int \frac{x^2-x^{-2}}{\sqrt{x^2+x^{-2}+1}} \frac{d(x-x^{-1})}{x+x^{-1}} \\
&= \int \frac{(x-x^{-1}) \, d(x - x^{-1})}{\sqrt{(x-x^{-1})^2+3}} \\
&= \int d\sqrt{(x-x^{-1})^2+3} \\
&= \sqrt{x^2+x^{-2}+1} + \text{const.}
\end{align}$$
