Evaluation of $\int{\dfrac{dx}{1+x^8}}$

Question

$\int{\dfrac{dx}{1+x^8}}$

Any hint for solving this indefnite integral appreciated.

Answer 1

Maybe you can take the following substitution $$x=(I)^{1/4} y,$$ where $$I=\sqrt{-1}.$$ Then $$ \int \frac{1}{1+x^8}=\int\frac{I^{1/4}}{1-y^8}=\frac{I^{1/4}}{2}\int\frac{1}{1-y^4}+\frac{1}{1+y^4};$$ and $$\int \frac{1}{1-y^4}=\frac{1}{2}\int \frac{1}{1-y²}+\frac{1}{1+y^2}=\frac{1}{4}\log\frac{1+y}{1-y}+ \frac{1}{2}\arctan y;$$ as for $\frac{1}{1+y^4}$, you can take the substitution $y=I^{1/2}t$,then $$\int \frac{1}{1+y^4}=\frac{I^{1/2}}{2}\int\frac{1}{1-t^2}+\frac{1}{1+t^2}=\frac{I^{1/2}}{4}\log\frac{1+t}{1-t}+\frac{I^{1/2}}{2}\arctan t;$$ hence we have $$\int\frac{1}{1+x^8}=\frac{I^{1/4}}{8}\log\frac{1+y}{1-y}+\frac{I^{1/4}}{4}\arctan y+\frac{I^{3/4}}{8}\log\frac{1+t}{1-t}+\frac{I^{3/4}}{4}\arctan t+c, $$ where $c$ is a constant and $y=I^{-1/4}x, t=I^{-3/4}x$.