How do you write the partial fraction decomposition of the rational expression  x/ (x^3-2x+x)?

Dec 14, 2015

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

Explanation:

Simplify: $\frac{x}{{x}^{3} - x}$

Factor the denominator.

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

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

If $x = 0$:

$0 = - A$
$A = 0$

If $x = 1$:

$1 = 2 C$
$C = \frac{1}{2}$

If $x = - 1$:

$- 1 = 2 B$
$B = - \frac{1}{2}$

Therefore, the expression is equal to

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