# How do I decompose the rational expression (x-3)/(x^3+3x) into partial fractions?

Jun 17, 2018

The answer is $= - \frac{1}{x} + \frac{x + 1}{{x}^{2} + 3}$

#### Explanation:

Perform the decomposition into partial fractions after factorising the denominator

$\frac{x - 3}{{x}^{3} + 3 x} = \frac{x - 3}{x \left({x}^{2} + 3\right)}$

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

$= \frac{A \left({x}^{2} + 3\right) + x \left(B x + C\right)}{x \left({x}^{2} + 3\right)}$

The denominators are the same, compare the numerators

$x - 3 = A \left({x}^{2} + 3\right) + x \left(B x + C\right)$

Let $x = 0$, $\implies$, $- 3 = 3 A$, $\implies$, $A = - 1$

Coefficients of ${x}^{2}$

$0 = A + B$, $\implies$, $B = - A = 1$

Coefficients of $x$

$1 = C$

Therefore,

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