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.

$$I=\int\frac{a^4-x^4}{x^4+a^2x^2+a^4}\,dx=a\int\frac{1-y^4}{y^4+y^2+1}\,dy=a\int\frac{1-y^4}{\left(y^2-y+1\right) \left(y^2+y+1\right) }\,dy$$
$$I=\frac a2\int\Bigg[\frac{\left(y+\frac 12\right)+\frac 32}{y^2+y+1}-\frac{\left(y-\frac 12\right)-\frac 32}{y^2-y+1}-1 \Bigg]\,dy$$ So, beside the linear term (and the integration constant), you should end with a logarithm and an arctangent.
