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

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

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

Hint: Use that $$\frac{x^4+1}{x^6-1}=\frac16\,{\frac {-2\,x-1}{{x}^{2}+x+1}}-\frac13\, \left( x+1 \right) ^{-1}+\frac16 \,{\frac {2\,x-1}{{x}^{2}-x+1}}+\frac13\, \left( x-1 \right) ^{-1}. $$

  • $\begingroup$ thanks..1/3 is easy to find..how to go about others? should i take lcm. $\endgroup$ – J. Deff Oct 14 '18 at 13:55
  • $\begingroup$ @J.Deff: which is of the form $$\frac{f'(x)}{f(x)}$$, so easy! $\endgroup$ – Chinnapparaj R Oct 14 '18 at 13:56
  • $\begingroup$ Yes. But how to find f'(x) using easy method because when i take lcm it becomes too long $\endgroup$ – J. Deff Oct 14 '18 at 13:57
  • $\begingroup$ Use that $$(x^2+x+1)'=2x+1$$ $\endgroup$ – Dr. Sonnhard Graubner Oct 14 '18 at 13:58
  • 1
    $\begingroup$ Your result looks like that $$-1/6\,\ln \left( {x}^{2}+x+1 \right) +1/3\,\ln \left( x-1 \right) +1 /6\,\ln \left( {x}^{2}-x+1 \right) -1/3\,\ln \left( x+1 \right) $$ $\endgroup$ – 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: $$\frac{x^4}{(x-1)(x+1)(x^2-x+1)(x^2+x+1)}$$


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