Re-write the denominator as x*(x^6-1)=x*(x^3-1)*(x^3+1)

[hadi amiri 4]

Can anyone help me with this \int \frac{dx}{x^7-x}?

 

[arildno]

1. Re-write the denominator as x*(x^6-1)=x*(x^3-1)*(x^3+1)

2. Use polynomial division to reduce the third-degree polynomials:
(x^{3}\pm{1}):(x\pm{1})=x^{2}\mp{x}+1

3. See if these can be factorized any further, then use partial fractions decomposition.

 

[hadi amiri 4]

but it takes an hour to do it
is there any simpler method for this?

 

[HallsofIvy]

I guess you had better stick to trivial problems!

 

[saeed69]

1/(x^7-x) = 1/(x*(x^6-1))= -1/x+x^5/(x^6-1)
int(-1/x+x^5/(x^6-1), x)=
int(-1/x, x)=-ln(x)
int(x^5/(x^6-1), x)=(1/6)*ln(x^6-1)
int(1/(x^7-x), x)=-ln(x)+(1/6)*ln(x^6-1)

 

[arildno]

1/(x^7-x) = 1/(x*(x^6-1))= -1/x+x^5/(x^6-1)

I didn't see that one..

Very good, saeed69!