Question

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

Answer 1

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

Explanation:

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"`

Answer 2

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

Explanation:

`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`