Partial fraction of $\frac 1{x^6+1}$ Can someone please help me find the partial fraction of $$1\over{x^6+1}$$
?
I know the general method of how to find the partial fraction of functions but this seems a special case to me..

We have
$$\begin{align}x^6+1&=(x^2)^3+1\\&=(x^2+1)(x^4-x^2+1)\\&=(x^2+1)(x^4+2x^2+1-3x^2)\\&=(x^2+1)((x^2+1)^2-(\sqrt 3\ x)^2)\\&=(x^2+1)(x^2-\sqrt 3\ x+1)(x^2+\sqrt 3\ x+1)\end{align}$$
Now setting $$\frac{1}{x^6+1}=\frac{a}{x^2+1}+\frac{bx^2+c}{x^4-x^2+1}$$
will give you $a=\frac 13,b=-\frac 13,c=\frac 23.$
Then, set 
$$\frac{-\frac 13x^2+\frac 23}{x^4-x^2+1}=\frac{dx+e}{x^2-\sqrt 3\ x+1}+\frac{fx+g}{x^2+\sqrt 3\ x+1}$$
to get $d,e,f,g$.
