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

Mar 10, 2018

$\int \frac{1}{{x}^{8} - x} \mathrm{dx} = \frac{1}{7} \ln \left\mid \frac{{x}^{7} - 1}{x} ^ 7 \right\mid + \text{c}$

#### Explanation:

We want to find $\int \frac{1}{{x}^{8} - x} \mathrm{dx}$.

We start by transforming the integrand into something more integrable.

$\frac{1}{{x}^{8} - x} = \frac{1}{\left(1 - {x}^{-} 7\right) {x}^{8}} = {x}^{-} \frac{8}{1 - {x}^{-} 7} = \frac{7 {x}^{-} 8}{7 \left(1 - {x}^{-} 7\right)}$

So

$\int \frac{1}{{x}^{8} - x} \mathrm{dx} = \int \frac{7 {x}^{-} 8}{7 \left(1 - {x}^{-} 7\right)} \mathrm{dx}$

Now let $u = - {x}^{-} 7$ and $\mathrm{du} = 7 {x}^{-} 8$ and substitute this into the integral

$\int \frac{7 {x}^{-} 8}{7 \left(1 - {x}^{-} 7\right)} \mathrm{dx} = \frac{1}{7} \int \frac{1}{1 + u} \mathrm{du} = \frac{1}{7} \ln \left\mid 1 + u \right\mid + \text{c}$

Now substitute back for $x$

$\frac{1}{7} \ln \left\mid 1 + u \right\mid + \text{c"=1/7lnabs(1-x^-7)+"c"=1/7lnabs((x^7-1)/x^7)+"c}$

Mar 10, 2018

$\int \frac{1}{{x}^{8} - x} \mathrm{dx} = \frac{1}{7} \ln \left\mid {x}^{7} - 1 \right\mid - \ln \left\mid x \right\mid + C$

#### Explanation:

$\frac{1}{{x}^{8} - x} = \frac{1}{x \left({x}^{7} - 1\right)}$

$\textcolor{w h i t e}{\frac{1}{{x}^{8} - x}} = \frac{A}{x} + \frac{B {x}^{6} + C {x}^{5} + D {x}^{4} + E {x}^{3} + F {x}^{2} + G x + H}{{x}^{7} - 1}$

Multiplying both ends by ${x}^{8} - x$ we get:

$1 = A \left({x}^{7} - 1\right) + \left(B {x}^{6} + C {x}^{5} + D {x}^{4} + E {x}^{3} + F {x}^{2} + G x + H\right) x$

Hence:

$A = - 1$

$B = 1$

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

So:

$\int \frac{1}{{x}^{8} - x} \mathrm{dx} = \int {x}^{6} / \left({x}^{7} - 1\right) - \frac{1}{x} \mathrm{dx}$

$\textcolor{w h i t e}{\int \frac{1}{{x}^{8} - x} \mathrm{dx}} = \frac{1}{7} \ln \left\mid {x}^{7} - 1 \right\mid - \ln \left\mid x \right\mid + C$