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?

$$\int\frac{xdx}{x^2-x+1}=\int\frac{(x-\frac12)dx}{x^2-x+1}+\int\frac{\frac12dx}{(x-\frac12)^2+\frac34}$$.  I'm guessing you know how to solve that.  The other half is very similar.
