Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How should I proceed about this integral? $$\int {1/(x^7 -x)} dx$$

I've tried integration by parts or substitution but I can't seem to solve it. Can I have some hints on how should I get started?

These are some of the things I've tried:

IBP: $u = \frac {1}{x^6-1}$, $du = \frac {-5x^6}{x^6-1}$, $dv = \frac 1x dx$, $v = \ln|x|$

Tried substitution method, but not successful.

share|cite|improve this question
If you can't think of anything else: $x^7-x=x(x^3+1)(x^3-1)=x(x+1)(x^2-x+1)(x-1)(x^2+x+1)$ and use partial fractions. – Mark Bennet Nov 14 '12 at 21:07
For the lazy people who don't want to write out the partial fractions: WolframAlpha. – TMM Nov 14 '12 at 21:15
up vote 10 down vote accepted

The cheater's method is to "observe" that $$ \frac{d}{dx} \ln\left(1-\frac{1}{x^6}\right) = \frac{6}{x^7-x} $$

share|cite|improve this answer
I wouldn't say it's cheater's method. Great intuition. – Kaster Nov 14 '12 at 21:29
How to get to this? - Well looking at taking "partial" partial fractions with factors $x(x^6-1)$ - with the intuition that there might be something going on with the sixth roots of unity seems to get there - [$\ln(x^6-1)-6\ln x$] – Mark Bennet Nov 14 '12 at 21:39
Thank you @GEdgar for you explanation. Im kinda get what you trying to say. you provided me with the fastest way to solve this question :D – melyong Nov 15 '12 at 5:10

One way to look at the problem is to say that it would be easy if the integrand were $$\frac{7x^6-1}{x^7-x},$$and also easy if it were $$\frac{x^6-1}{x^7-x}.$$Now take a linear combination of these to knock out the $x^6$ term in the numerator.

share|cite|improve this answer

Partial fractions: $x^7-x=x(x^6-1)=x(x^3-1)(x^3+1)$, and then use the standard factorizations of $a^3-b^3$ and $a^3+b^3$ to split each of the cubic factors into a linear factor and a quadratic factor.

share|cite|improve this answer

As in this example it is useful to extract something of the form $$u=x^a\pm\frac{1}{x^a}$$ $$\int\frac{dx}{x^{7}-x}=\int\frac{1}{x^{4}}\frac{dx}{x^{3}-\frac{1}{x^{3}}}=\int\frac{\left(x^{2}+\frac{1}{x^{4}}\right)dx}{x^{3}-\frac{1}{x^{3}}}-\int\frac{x^{2}dx}{x^{3}-\frac{1}{x^{3}}}=I_1-I_2$$ Now in $I_1$ we may let $$u=x^{3}-\frac{1}{x^{3}}$$ $$du=3\left(x^{2}+\frac{1}{x^{4}}\right)dx$$ So that

$$I_1=\frac{1}{3}\int\frac{du}{u}=\frac{1}{3}\ln|u|=\frac{1}{6}\ln u^2=\frac{1}{6}\ln\frac{(x^6-1)^2}{x^6}$$ In $I_2$ apparently $v=x^3$




share|cite|improve this answer
thank you so much for explaining the approach in great details. May I know what is the name of this method as I did not learn this technique of integration in school. – melyong Nov 15 '12 at 5:01
I haven't seen a systematic discussion of this method and just inferred it from several examples. I am also not convinced whether it works fine unless for the carefully crafted textbook excercises – Valentin Nov 15 '12 at 9:02
I see... I still have to thank for teaching me this new method to solve this particular question. – melyong Nov 15 '12 at 11:23

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.