# Integrate $\int \frac{x^4 +1}{x^6 - 1}dx$

Integrate

$$\int \frac{x^4 +1}{x^6 - 1}\, \mathrm dx$$

I have tried using partial fractions but to no use. Thanks for help.

• hint: $x^6-1=(x-1)(x+1)(x^2-x+1)(x^2+x+1)$. – thesmallprint Oct 14 '18 at 13:53

Hint: Use that $$\frac{x^4+1}{x^6-1}=\frac16\,{\frac {-2\,x-1}{{x}^{2}+x+1}}-\frac13\, \left( x+1 \right) ^{-1}+\frac16 \,{\frac {2\,x-1}{{x}^{2}-x+1}}+\frac13\, \left( x-1 \right) ^{-1}.$$
• @J.Deff: which is of the form $$\frac{f'(x)}{f(x)}$$, so easy! – Chinnapparaj R Oct 14 '18 at 13:56
• Use that $$(x^2+x+1)'=2x+1$$ – Dr. Sonnhard Graubner Oct 14 '18 at 13:58
• Your result looks like that $$-1/6\,\ln \left( {x}^{2}+x+1 \right) +1/3\,\ln \left( x-1 \right) +1 /6\,\ln \left( {x}^{2}-x+1 \right) -1/3\,\ln \left( x+1 \right)$$ – Dr. Sonnhard Graubner Oct 14 '18 at 14:01
I believe that partial fractions is the easiest way to do this. Note: $$(x^6-1)=(x-1)(x+1)(x^2-x+1)(x^2+x+1)$$ so you will want to do the partial fractions of: $$\frac{x^4}{(x-1)(x+1)(x^2-x+1)(x^2+x+1)}$$