Decomposition into partial fractions to compute an integral I'm having problems with: 
$$\int_{-\infty}^{\infty}\frac{x^4+1}{x^6+1}dx$$
I was thinking: $\frac{x^4+1}{x^6+1}$ is an even function and the interval $(-\infty,\infty)$ is symmetric about 0, we could write the integral like:
$$2\int_{0}^{\infty}\frac{(x^4+1)}{x^6+1}dx$$
And for $\frac{x^4+1}{x^6+1}$, I will use the partial fractions. I will write $x^6+1$ like a sum of cubes $x^6+1=(x^2)^3+1^3$ and use the formula $a^3+b^3=(a+b)(a^2-ab+b^2)$ and in the end we will have 
$$x^6+1=(x^2+1)(x^4-x^2+1)$$ But now I'm stuck and do not know how to decompose $x^4-x^2+1$. I was thinking about $(x^2-1)x^2+1$ or $(x^2-\frac{1}{2})^2+\frac{3}{4}$ but I need a form for that like $-(-1+\sqrt3 x-x^2) (1+\sqrt3 x+x^2)$...
A little help here?

HINT:
As $y^3+1=(y+1)(y^2-y+1),$
$$\dfrac{x^4+1}{x^6+1}=\dfrac{x^4-x^2+1}{x^6+1}+\dfrac{x^2}{x^6+1}=\dfrac1{x^2+1}+\dfrac{x^2}{x^6+1}$$
Set $x^3=u$ for the second part
