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

Any hint for solving this indefnite integral appreciated.

HeroZhang001
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M.R. Yegan
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  • If you are integrating from $-\infty$ to $\infty$, choosing a convenient contour would be the quickest way in my opinion. If not, the computation should not be quick – Adrien Martina Mar 06 '23 at 13:59
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    Why would you guess that, @AdrienMartina? This is the usual format for an indefinite integral, and the question specifically asks for an indefinite integral. – Thomas Andrews Mar 06 '23 at 14:00
  • Based on this result, you can factor $x^8+1$ using roots of unity, use partial fractions, and integrate – Тyma Gaidash Mar 06 '23 at 14:00

1 Answers1

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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 $y=I^{-1/4}x, t=I^{-3/4}x$.

Dennis Jia
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