Calculating $\int\frac{a^4-x^4}{x^4+a^2x^2+a^4}dx$ 
$$\int\frac{a^4-x^4}{x^4+a^2x^2+a^4}dx$$

I tried factoring as $${(a^2-x^2)(a^2+x^2)}$$ $$\text{and}$$ $$(x^2+a^2+ax)(x^2+a^2-ax)$$ but didn't get much after this. I sew that the expression is symmetric in a,x but I don't know how to use that.

Hint:
First,
$$\dfrac{a^4 - x^4}{x^4 + a^2x^2 + a^4} = \dfrac{a^2x^2 + 2a^4}{x^4 + a^2x^2 + a^4} - 1 = a^2\dfrac{x^2 + 2a^2}{x^4 + a^2x^2 + a^4} - 1$$
Second,
$$\dfrac{x^2 + 2a^2}{x^4 + a^2x^2 + a^4} = \dfrac{x^2 + 2a^2}{\left(x^2 -ax + a^2\right)\left(x^2 + ax + a^2\right)}$$
Now, perform partial-fraction decomposition.
