<s>[INST] 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..
 [/INST] \begin{align}
\frac 1{x^6+1}&=\frac 1{(x^2)^3+1}\\
&=\frac 1{((x^2)1+)((x^2)^2-(x^2)+1)} \\
&= \frac 1{(x^2+1)(x^4-x^2+1)}  \\
&= \frac {Ax+B}{x^2+1}+\frac{Cx+D}{x^4-x^2+1}
\end{align}
So $$1=(x^4-x^2+1)(Ax+B)+(x^2+1)(Cx+D)$$
At $x=0$, the equation becomes $$1=B+D$$
At this point, equating the coefficients is the only way to go.
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