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Evaluate the integral of (1 – x3) / (x(x2 + 1))

by RoRi

Compute the following integral.

    \[ \int \frac{1-x^3}{x(x^2+1)} \, dx. \]

First, we factor the numerator and do some rearranging of the integral,

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

For the remaining integral \int \frac{1}{x(x^2+1)} \, dx we use partial fractions, writing

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

This gives us the equation

    \[ A(x^2+1) + (Bx+C)x = 1. \]

First, we evaluate at x = 0 to obtain

    \[ A = 1. \]

Then we have

    \[ x^2+1 + Bx^2 + Cx = 1 \quad \implies \quad (1+B)x^2 + Cx = 0. \]

Equating like powers of x we have B = -1 and C = 0. Therefore,

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

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