Find $\int \frac{1}{x^4+x^2+1} \,\, dx$ Find $$\int \frac{1}{x^4+x^2+1} \,\, dx$$

I tried to find like that:
$\int \frac{1}{x^4+x^2+1} = \int \frac{\frac{1}{2}x + \frac{1}{2}}{x^2+x+1} \,\, dx + \int \frac{-\frac{1}{2}x + \frac{1}{2}}{x^2-x+1} \,\, dx = \frac{1}{2} \Big(\int \frac{2x + 1}{x^2+x+1} - \int \frac{x}{x^2+x+1} \Big) - \frac{1}{2} \Big( \int \frac{2x - 1}{x^2-x+1} \,\, dx - \int \frac{x}{x^2-x+1} \,\, dx \Big)$
but then I don't know how to find integrals:
$\, \int \frac{x}{x^2-x+1} \,\, dx \,$ and  $\, \int \frac{x}{x^2+x+1} \,\, dx \,$
Is there another way to integrate this function or way to end my calculations?

simple methods,Note
$$1=\dfrac{1}{2}(x^2+1)+\dfrac{1}{2}(1-x^2)$$
so
$$\int\dfrac{1}{x^4+x^2+1}dx=\frac{1}{2}(I_{1}+I_{2})$$
$$I_{1}=\int\dfrac{x^2+1}{x^4+x^2+1}dx=\int\dfrac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}+1}dx=\int\dfrac{d(x-\frac{1}{x})}{(x-1/x)^2+3}$$
and simaler $I_{2}$
