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Evaluate the integral of 1 / (x3 – x)

by RoRi

Compute the following integral.

    \[ \int \frac{dx}{x^3-x}. \]

Since x^3 - x factors as x(x^2-1) = x(x+1)(x-1) we have the following,

    \[ \frac{1}{x^3 - x} = \frac{A}{x} + \frac{B}{x+1} + \frac{C}{x-1}. \]

This gives us the equation

    \[ A (x+1)(x-1) + B(x)(x-1) + C(x)(x+1) = 1. \]

Substituting the values x = 0, x = -1 and x = 1 we obtain

    \begin{align*} -A &= 1 & \implies \qquad A &= -1 \\ 2B &= 1 & \implies \qquad B &= \frac{1}{2} \\ 2C &= 1 & \implies \qquad C &= \frac{1}{2}. \end{align*}

Therefore, we have

    \begin{align*} \int \frac{dx}{x^3-x} &= \int \frac{-1}{x} \, dx + \frac{1}{2} \int \frac{1}{x+1} \, dx + \frac{1}{2} \int \frac{1}{x-1} \, dx \\ &= -\log|x| + \frac{1}{2} \log|x+1| + \frac{1}{2} \log|x-1| + C \\ &= \frac{1}{2} \log |x^2-1| - \log |x| + C. \end{align*}

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