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Evaluate the integral of (x+1) / (x3-1)

by RoRi

Compute the following integral.

    \[ \int \frac{x+1}{x^3-1} \, dx. \]

Since the denominator factors as

    \[ x^3 - 1 = (x-1)(x^2 + x + 1) \]

we use partial fractions and write

    \[ \frac{x+1}{x^3-1} = \frac{x+1}{(x-1)(x^2+x+1)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+x+1}. \]

This gives us the equation

    \[ A(x^2+x+1) + (Bx+C)(x-1) = x+1. \]

First, we evaluate at x = 1 to find the value of A,

    \[ 3A = 2 \quad \implies \quad A = \frac{2}{3}. \]

Then, using this value of A and evaluating at x = 0 to obtain the value of C,

    \[ \frac{2}{3} - C = 1 \qquad \implies \qquad C = -\frac{1}{3}. \]

Finally, we evaluate at x = 2 and use the values of A and C to compute B,

    \[ \frac{14}{3} + 2B -\frac{1}{3} = 3 \qquad \implies \qquad B = -\frac{2}{3}. \]

Therefore,

    \begin{align*} \int \frac{x+1}{x^3-1} \, dx &= \int \frac{x+1}{(x-1)(x^2+x+1)} \, dx \\[9pt] &= \int \left( \frac{A}{x-1} + \frac{Bx+C}{x^2+x+1} \right) \, dx \\[9pt] &= \frac{2}{3} \int \frac{1}{x-1} \, dx - \frac{1}{3} \int \frac{2x+1}{x^2+x+1} \, dx \\[9pt] &= \frac{2}{3} \log |x-1| - \frac{1}{3} \log(x^2+x+1) + C \\[9pt] &= \frac{1}{3} \log \left(\frac{(x-1)^2}{x^2+x+1}\right) + C. \end{align*}

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