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Evaluate the integral of (x-3) / (x3 + 3x2 + 2x)

by RoRi

Compute the following integral.

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

The denominator of the integrand factors as

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

Therefore, we can use partial fractions as follows,

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

This gives us the equation

    \[ A(x+1)(x+2) + B(x)(x+2) + C(x)(x+1) = x-3. \]

Evaluating at x = 0, x = -1, and x = -2 we obtain

    \begin{align*} 2A &= -3 & \implies \qquad A &= -\frac{3}{2} \\ -B &= -4 & \implies \qquad B &= 4 \\ 2C &= -5 & \implies \qquad C &= -\frac{5}{2}. \end{align*}

Therefore, we have

    \begin{align*} \int \frac{x-3}{x^3 + 3x^2 + 2x} \, dx &= \int \left( \frac{A}{x} + \frac{B}{x+1} + \frac{C}{x+2} \right) \, dx \\[10pt] &= -\frac{3}{2} \int \frac{1}{x} \, dx + 4 \int \frac{1}{x+1} \, dx - \frac{5}{2} \int \frac{1}{x+2} \, dx \\[10pt] &= -\frac{3}{2} \log |x| + 4 \log |x+1| - \frac{5}{2} \log |x+2| + C \\[10pt] &= 4 \log |x+1| - \frac{3}{2} \log |x| - \frac{5}{2} \log |x+2| + C. \end{align*}

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