# How do you write the partial fraction decomposition of the rational expression  (x-1)/(x^3 +x)?

Oct 29, 2016

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

#### Explanation:

Let's factorise the denominator ${x}^{3} + x = x \left({x}^{2} + 1\right)$
so the partial fraction decomposition is
$\frac{x - 1}{{x}^{3} + x} = \frac{x - 1}{x \left({x}^{2} + 1\right)} = \frac{A x + B}{{x}^{2} + 1} + \frac{C}{x}$
$= \frac{x \left(A x + B\right) + C \left({x}^{2} + 1\right)}{\left(x\right) \left({x}^{2} + 1\right)}$
So we have $x - 1 = x \left(A x + B\right) + C \left({x}^{2} + 1\right)$
let $x = 0$$\implies$$- 1 = C$$\implies$$C = - 1$
Coefficient of x, $1 = B$$\implies$$B = 1$
coefficients of x^2, $0 = A + C$ $\implies$$A = 1$

And finally we have
$\frac{x - 1}{{x}^{3} + x} = \frac{x + 1}{{x}^{2} + 1} - \frac{1}{x}$