# How do you write the partial fraction decomposition of the rational expression x/((x-1)(x^2+4)?

Nov 11, 2016

The answer is $= \frac{\frac{1}{5}}{x - 1} + \frac{- \frac{1}{5} x + \frac{4}{5}}{{x}^{2} + 4}$

#### Explanation:

Let's write the decomposition
$\frac{x}{\left(x - 1\right) \left({x}^{2} + 4\right)} = \frac{A}{x - 1} + \frac{B x + C}{{x}^{2} + 4}$

$= \frac{A \left({x}^{2} + 4\right) + \left(B x + C\right) \left(x - 1\right)}{\left(x - 1\right) \left({x}^{2} + 4\right)}$

$\therefore x = A \left({x}^{2} + 4\right) + \left(B x + C\right) \left(x - 1\right)$

Let $x = 1$$\implies$$1 = 5 A$$\implies$$A = \frac{1}{5}$
Coefficients of ${x}^{2}$$\implies$$0 = A + B$$\implies$ $B = - \frac{1}{5}$
And $0 = 4 A - C$$\implies$$C = \frac{4}{5}$

So, $\frac{x}{\left(x - 1\right) \left({x}^{2} + 4\right)} = \frac{\frac{1}{5}}{x - 1} + \frac{- \frac{1}{5} x + \frac{4}{5}}{{x}^{2} + 4}$