How can i integrate #1/(x^8-x)#?

Question

Isabella Ackerman · Accepted Answer

#int1/(x^8-x) dx=1/7lnabs((x^7-1)/x^7)+"c"#

We want to find #int1/(x^8-x)dx#.

We start by transforming the integrand into something more integrable.

#1/(x^8-x)=1/((1-x^-7)x^8)=x^-8/(1-x^-7)=(7x^-8)/(7(1-x^-7))#

So

#int1/(x^8-x)dx=int(7x^-8)/(7(1-x^-7))dx#

Now let #u=-x^-7# and #du=7x^-8# and substitute this into the integral

#int(7x^-8)/(7(1-x^-7))dx=1/7int1/(1+u)du=1/7lnabs(1+u)+"c"#

Now substitute back for #x#

#1/7lnabs(1+u)+"c"=1/7lnabs(1-x^-7)+"c"=1/7lnabs((x^7-1)/x^7)+"c"#

Isabella Ackerman · Answer

#int 1/(x^8-x) dx = 1/7 ln abs(x^7-1)- ln abs(x) + C#

#1/(x^8-x) = 1/(x(x^7-1))#

#color(white)(1/(x^8-x)) = A/x + (Bx^6+Cx^5+Dx^4+Ex^3+Fx^2+Gx+H)/(x^7-1)#

Multiplying both ends by #x^8-x# we get:

#1 = A(x^7-1) + (Bx^6+Cx^5+Dx^4+Ex^3+Fx^2+Gx+H)x#

Hence:

#A = -1#

#B = 1#

#C = D = E = F = G = H = 0#

So:

#int 1/(x^8-x) dx = int x^6/(x^7-1)-1/x dx#

#color(white)(int 1/(x^8-x) dx) = 1/7 ln abs(x^7-1)- ln abs(x) + C#

Scarlett Allison · Answer

undefined To integrate $ \frac{1}{x^8 - x} $, you can use partial fraction decomposition followed by integration. The first step is to factor the denominator, which is $ x(x^7 - 1) $. Then, you decompose $ \frac{1}{x^8 - x} $ into partial fractions. After decomposing, integrate each term separately.

The decomposition will have the form:

$$ \frac{1}{x(x^7 - 1)} = \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{x - \omega} + \frac{D}{x - \omega^2} + \ldots + \frac{H}{x - \omega^7} $$

Where $ \omega $ is a complex number, a primitive 8th root of unity.

Once you find the constants $ A, B, C, \ldots, H $, integrate each term separately. This integration might involve logarithms and other elementary functions depending on the specific values of the constants.

However, note that integrating rational functions with high-degree polynomials in the denominator can be complex, and the resulting integrals may not have elementary solutions in terms of standard functions.