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

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Answer 1 · 2018-03-10T15:49:33

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

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 $x$

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

Answer 2 · 2018-03-10T15:59:14

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

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

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

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:

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

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

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