# how to integrate $\int 1/(x^7 -x) dx$??

I serious have no idea how should i proceed about doing this question. I've tried IBP or substitution but I can't seems to solve it. Can I have some hints of how should I get started?

$\int {1/(x^7 -x)} dx$

These are some of the stuffs 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..

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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

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

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 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

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.

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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.

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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 $v=x^3$

$$I_2=\frac{1}{3}\int\frac{vdv}{v^{2}-1}=\frac{1}{6}\ln\left(v^{2}-1\right)=\frac{1}{6}\ln|x^6-1|$$

Finally,

$$I=I_1-I_2=\frac{1}{6}\ln\frac{x^6-1}{x^6}=\frac{1}{6}\ln\left(1-\frac{1}{x^6}\right)$$

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 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