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Evaluate the integral of x / (x3 – 3x + 2)

by RoRi

Compute the following integral.

    \[ \int \frac{x \, dx}{x^3 - 3x + 2}. \]

First, we factor the denominator,

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

Then we use partial fractions to decompose the integrand,

    \begin{align*} \frac{x}{x^3-3x+2} &= \frac{x}{(x-1)^2(x+2)} \\ &= \frac{A}{x+2} + \frac{B}{x-1} + \frac{C}{(x-1)^2}. \end{align*}

This gives us the equation

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

Substituting the values x = 1 and x = -2

    \begin{align*} 3C &= 1 &\implies \qquad C &= \frac{1}{3} \\ 9A &= -2 & \implies \qquad A &= -\frac{2}{9}. \end{align*}

Then we use these values of A and C and plug in x = 0 to find B,

    \begin{align*} A + -2B + 2C = 0 && \implies && 2B &= A + 2C \\ && \implies && 2B &= -\frac{2}{9} + \frac{2}{3} \\ && \implies && B &= \frac{2}{9}. \end{align*}

So, now we can evaluate the integral with this partial fraction decomposition,

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

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