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Evaluate the integral of 1 / (x4 – 1)

by RoRi

Compute the following integral.

    \[ \int \frac{dx}{x^4-1}. \]

First, we factor the denominator

    \[ x^4 - 1= (x^2 - 1)(x^2+1) = (x-1)(x+1)(x^2+1). \]

To apply partial fraction decomposition we then write,

    \[ \frac{1}{x^4-1} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{Cx+D}{x^2+1}. \]

This gives us the equation

    \[ A(x+1)(x^2+1) + B(x-1)(x^2+1) + (Cx+D)(x-1)(x+1) = 1. \]

First, we evaluate the equation at x = 1 and x = -1 to obtain

    \begin{align*} 4A &= 1 & \implies \qquad A &= \frac{1}{4} \\ -4B &= 1 & \implies \qquad B &= -\frac{1}{4}. \end{align*}

Then, using these values of A and B we evaluate at x = 0 to obtain

    \[ \frac{1}{4} + \frac{1}{4} - D = 1 \qquad \implies \qquad D = -\frac{1}{2}. \]

Finally, equating like powers of x in the equation we must have C = 0. Therefore,

    \begin{align*} \int \frac{1}{x^4-1} \, dx &= \int \left( \frac{A}{x-1} + \frac{B}{x+1} + \frac{Cx+D}{x^2+1} \right) \, dx \\[9pt] &= \frac{1}{4} \int \frac{1}{x-1} \, dx - \frac{1}{4} \int \frac{1}{x+1} \, dx - \frac{1}{2} \int \frac{1}{x^2+1} \, dx \\[9pt] &= \frac{1}{4} \log |x-1| - \frac{1}{4} \log |x+1| - \frac{1}{2} \arctan x + C\\[9pt] &= \frac{1}{4} \log \left| \frac{x-1}{x+1} \right| - \frac{1}{2} \arctan x + C. \end{align*}

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