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

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Answer 1 · 2015-12-14T05:32:35

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

Simplify: $\frac{x}{x^{3} - x}$ $x/(x^3-x)$

Factor the denominator.

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

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

If $x = 0$ $x=0$ :

$0 = - A$ $0=-A$
$A = 0$ $A=0$

If $x = 1$ $x=1$ :

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

If $x = - 1$ $x=-1$ :

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

Therefore, the expression is equal to

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

How do you write the partial fraction decomposition of the rational expression x/ (x^3-2x+x)xx3−2x+x x/ (x^3-2x+x)?