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?
Integrate $\int \frac{1}{x^6-1}dx$
indefinite-integralsintegration
Best Answer
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}