# How to integrate $\int \frac{1}{x^6-1}dx$

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?

• Commented Apr 25, 2023 at 13:59
• What about partial fraction decomposition over complex numbers? Commented Apr 25, 2023 at 14:48

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}
• 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 $\int \frac{1}{x^6-1}dx=\frac12\int \frac1{x^3-1}-\frac1{x^3+1}\ dx$ Commented Apr 25, 2023 at 15:05
• Something's wrong... $\frac1{x^3-1}-\frac1{x^3-1}=0$ Commented Apr 25, 2023 at 15:05