How to integrate $$\int \frac{1}{x^6-1}dx$$ I have done it by using $$x^6-1=(x-1)(x+1)(x^2+x+1)(x^2-x+1)$$ and then applying partial fraction decomposition but this makes for a very long process.
Is there a better way to integrate this function?
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$\begingroup$ See art of problem solving. $\endgroup$– Dietrich BurdeCommented Apr 25, 2023 at 13:59
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3$\begingroup$ What about partial fraction decomposition over complex numbers? $\endgroup$– Bob DobbsCommented Apr 25, 2023 at 14:48
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1 Answer
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Decompose the integrand as $$\frac2{x^6-1}= \frac1{x^2-1}- \frac{x^2}{x^6-1}-\frac{x^2+1}{x^4+x^2+1} $$ and then integrate piecewise to obtain
\begin{align} &\int \frac{1}{x^6-1}dx\\ =&\ \frac12\int \frac1{x^2-1}- \frac{x^2}{x^6-1}-\frac{x^2+1}{x^4+x^2+1}\ dx\\ =&\ \frac12\int \frac{dx}{x^2-1}- \frac{\frac13 d(x^3)}{(x^3)^2-1}-\frac{d(x-\frac1x)}{(x-\frac1x)^2+3}\\ =&\ \frac14\ln\frac{x-1}{x+1}-\frac1{12}\ln\frac{x^3-1}{x^3+1}-\frac1{2\sqrt3}\tan^{-1}\frac{x-\frac1x}{\sqrt3} +C \end{align}
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$\begingroup$ There is a typo here $ \int \frac{1}{x^6-1}dx=\frac12\int \frac1{x^3-1}-\frac1{x^3-1}\ dx$- it should be $ \int \frac{1}{x^6-1}dx=\frac12\int \frac1{x^3-1}-\frac1{x^3+1}\ dx$ $\endgroup$ Commented Apr 25, 2023 at 15:05
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$\begingroup$ Something's wrong... $\frac1{x^3-1}-\frac1{x^3-1}=0$ $\endgroup$ Commented Apr 25, 2023 at 15:05
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