# How do you evaluate the integral int 1/(x^3-x)dx?

Mar 6, 2017

$- \ln \left\mid x \right\mid + \left(\frac{1}{2}\right) \ln \left\mid x - 1 \right\mid + \left(\frac{1}{2}\right) \ln \left\mid x + 1 \right\mid + C$

#### Explanation:

This can be solved by using the method of partial fractions.

$\int \frac{1}{{x}^{3} - x} \mathrm{dx}$

First factor the denominator completely.

$\int \frac{1}{x \left({x}^{2} - 1\right)} \mathrm{dx} = \int \frac{1}{x \left(x - 1\right) \left(x + 1\right)} \mathrm{dx}$

Decompose the integrand.

$\frac{1}{x \left(x + 1\right) \left(x - 1\right)} = \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{x + 1}$

Get the lowest common denominator between all of the terms of the decomposition.

$\frac{1}{x \left(x + 1\right) \left(x - 1\right)} = \frac{A \left(x - 1\right) \left(x + 1\right)}{x \left(x - 1\right) \left(x + 1\right)} + \frac{B x \left(x + 1\right)}{x \left(x - 1\right) \left(x + 1\right)} + \frac{C x \left(x - 1\right)}{x \left(x - 1\right) \left(x + 1\right)}$

Now that all terms have a common denominator, focus on the numerator to solve for the value of A, B, and C.

$1 = A \left(x - 1\right) \left(x + 1\right) + B x \left(x + 1\right) + C x \left(x - 1\right)$

For $x = 0$,
$1 = A \left(- 1\right) \left(1\right)$
$1 = - A$
$- 1 = A$

For $x = 1$,
$1 = B \left(1\right) \left(2\right)$
$1 = 2 B$
$\frac{1}{2} = B$

For $x = - 1$,
$1 = C \left(- 1\right) \left(- 2\right)$
$1 = 2 C$
$\frac{1}{2} = C$

Substitute the decomposition in to the integral.

$\int \frac{- 1}{x} + \frac{\frac{1}{2}}{x - 1} + \frac{\frac{1}{2}}{x + 1} \mathrm{dx}$

Break the integral into smaller integrals for each term

$\int \frac{- 1}{x} \mathrm{dx} + \int \frac{\frac{1}{2}}{x - 1} \mathrm{dx} + \int \frac{\frac{1}{2}}{x + 1} \mathrm{dx}$

Bring out the constants

$- \int \frac{1}{x} \mathrm{dx} + \frac{1}{2} \int \frac{1}{x - 1} \mathrm{dx} + \frac{1}{2} \int \frac{1}{x + 1} \mathrm{dx}$

Integrate each term

$- \ln \left\mid x \right\mid + \left(\frac{1}{2}\right) \ln \left\mid x - 1 \right\mid + \left(\frac{1}{2}\right) \ln \left\mid x + 1 \right\mid + C$