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Evaluate:

$$\int \frac{1}{x^7-x}\ \mathrm{d}x$$

My approach to this question:

$$\int \frac{1}{x^7-x}\ \mathrm{d}x = \int \frac{1}{x(x^6-1)}\ \mathrm{d}x$$ $$\int \frac{1}{x(x^6-1)}\ \mathrm{d}x = \int \frac{1}{x(x-1)(x+1)(x^2-x+1)(x^2+x+1)}\ \mathrm{d}x$$ $$\frac{1}{x(x-1)(x+1)(x^2-x+1)(x^2+x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1} + \frac{Dx+E}{x^2-x+1} + \frac{Fx+G}{x^2+x+1}$$

At this point I realized how brutal this question if going to be. Is there an easier way to solve the integral?

Ian Mateus
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Sc4r
  • 639

1 Answers1

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Yes there is. Note that: $$\frac{1}{x^7-x} =\frac{x^5}{x^6-1}-\frac{1}{x}$$ Now use $t = x^6-1$ for the first integral, and...

Note how this trick works for any integral of the form $(x^n-x)^{-1}$.

Nathaniel Bubis
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  • Thanks for this trick Nathaniel :) Solved it and got: $\frac16ln(x^6 -1)-ln(x) + C$ hopefully that's correct :) – Sc4r Dec 15 '13 at 23:32