How do you integrate (x-1)/(x^3 +x) using partial fractions?

Jan 8, 2017

Do a partial fraction expansion, check it, then integrate the terms.

Explanation:

Begin with the a partial fraction expansion:

$\frac{x - 1}{{x}^{3} + x} = \frac{A}{x} + \frac{B x + C}{{x}^{2} + 1}$

$x - 1 = A \left({x}^{2} + 1\right) + \left(B x + C\right) x$

$x - 1 = A \left({x}^{2} + 1\right) + B {x}^{2} + C x$

To make B and C disappear, let $x = 0$:

$0 - 1 = A \left({0}^{2} + 1\right)$

A = -1

$x + {x}^{2} = B {x}^{2} + C x$

$B = 1$
$C = 1$

Check:

$\frac{x}{{x}^{2} + 1} \frac{x}{x} + \frac{1}{{x}^{2} + 1} \frac{x}{x} - \frac{1}{x} \frac{{x}^{2} + 1}{{x}^{2} + 1}$

${x}^{2} / \left(x \left({x}^{2} + 1\right)\right) + \frac{x}{x \left({x}^{2} + 1\right)} - \frac{{x}^{2} + 1}{x \left({x}^{2} + 1\right)}$

$\frac{x - 1}{x \left({x}^{2} + 1\right)}$

$\frac{x - 1}{{x}^{3} + x}$

This checks.

$\int \frac{x - 1}{{x}^{3} + x} \mathrm{dx} = \int \frac{x}{{x}^{2} + 1} \mathrm{dx} + \int \frac{1}{{x}^{2} + 1} \mathrm{dx} - \int \frac{1}{x} \mathrm{dx}$

$\int \frac{x - 1}{{x}^{3} + x} \mathrm{dx} = \frac{1}{2} \ln \left({x}^{2} + 1\right) + {\tan}^{-} 1 \left(x\right) - \ln | x | + C$