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 :-(
On
I want to present another method such that we calculate 2 partial fractions incited of 4 partial fractions. By substitution $\frac{1}{x}=t$ we have $dx=-\frac{1}{t^2}dt$ and we obtain the form $J$ that is equal the form $I$ as follows: $$ \begin{array}{l} I=\int\,\frac{dx}{1+x^4+x^8} \\ \\ J=\int\, \frac{-\frac{1}{t^2}dt}{1+{(\frac{1}{t})}^4+{(\frac{1}{t})}^8} =J=\int\, \frac{-\frac{1}{t^2}dt}{\frac{t^8+t^4+1}{t^8}}= \int\,\frac{-t^6\,dt}{1+t^4+t^8} \end{array} $$ which results that $$ 2\,I=I+J=\int\,\frac{(1-t^6)\,dt}{1+t^4+t^8}= \int \, \frac{(1-t^2)(1+t^4+t^8)\,dt}{(t^4+t^2+1)(t^4-t^2+1)} =\int\, \frac{(1-t^2)\, dt}{(t^4-t^2+1)} $$
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.