<s>[INST] integral $\int_{}^{}\frac{dx}{1+x^4+x^8} $ looking for help for the following integral -
$$
\int_{}^{}\frac{dx}{1+x^4+x^8} 
$$
what I tried to do:
$$\int_{}^{}\frac{dx}{1+x^4+x^8}  = \int_{}^{}\frac{dx}{\frac14+x^4+x^8 + \frac{3}{4}}= \int_{}^{}\frac{dx}{\left(x^4+\frac{1}{2}\right)^2 + \frac{3}{4}} $$
and now I am stuck :-(
 [/INST] Factorise the denominator $x^8+x^4+1=(x^4+x^2+1)(x^4-x^2+1)=(x^2+x+1)(x^2-x+1)(x^2+\sqrt{3}x+1)(x^2-\sqrt{3}x+1)$. Now do partial fractions
\begin{eqnarray*}
\frac{1}{1+x^4+x^8}=\frac{1}{4(1+x+x^2)}+\frac{1}{4(1-x+x^2)}+\frac{2x+\sqrt{3}}{4\sqrt{3}(1+\sqrt{3}x+x^2)}-\frac{2x-\sqrt{3}}{4\sqrt{3}(1-\sqrt{3}x+x^2)}
\end{eqnarray*}
Each of these terms can be integrated using standard logs and inverse tangents.
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