# How do you express x/((x-1)(x^2+4) in partial fractions?

Nov 7, 2016

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

#### Explanation:

Let $\frac{x}{\left(x - 1\right) \left({x}^{2} + 4\right)} = \frac{A}{x - 1} + \frac{B x + C}{{x}^{2} + 4}$
$= \frac{A \left({x}^{2} + 4\right) + \left(B x + C\right) \left(x - 1\right)}{\left(x - 1\right) \left({x}^{2} + 4\right)}$
So $x = A \left({x}^{2} + 4\right) + \left(B x + C\right) \left(x - 1\right)$
if $x = 0$ $\implies$$0 = 4 A - C$
coefficients of ${x}^{2}$ $\implies$$0 = A + B$
coefficents of $x$ $\implies$ $1 = - B + C$
Solving for $A , B , C$
$A = \frac{1}{5}$
$B = - \frac{1}{5}$
$C = \frac{4}{5}$

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