# How do you express 10 / [(x-1)(x^2 + 9)] in partial fractions?

Sep 23, 2016

$\frac{10}{\left(x - 1\right) \left({x}^{2} + 9\right)} = \frac{1}{x - 1} - \frac{x + 1}{{x}^{2} + 9}$

#### Explanation:

Let $\frac{10}{\left(x - 1\right) \left({x}^{2} + 9\right)} \Leftrightarrow \frac{A}{x - 1} + \frac{B x + C}{{x}^{2} + 9}$.

Therefore $\frac{10}{\left(x - 1\right) \left({x}^{2} + 9\right)} \Leftrightarrow \frac{A \left({x}^{2} + 9\right) + \left(B x + C\right) \left(x - 1\right)}{\left(x - 1\right) \left({x}^{2} + 9\right)}$

$\Leftrightarrow \frac{A {x}^{2} + 9 A + B {x}^{2} - B x + C x - C}{\left(x - 1\right) \left({x}^{2} + 9\right)}$ or

$\frac{10}{\left(x - 1\right) \left({x}^{2} + 9\right)} \Leftrightarrow \frac{\left(A + B\right) {x}^{2} - \left(B - C\right) x + 9 A - C}{\left(x - 1\right) \left({x}^{2} + 9\right)}$

Hence $A + B = 0$, $B - C = 0$ and $9 A - C = 10$

or $A = - B = - C$ and so $9 \times \left(- C\right) - C = 10$ i.e. $- 10 C = 10$ i.e.

$C = - 1$, $A = 1$ and $B = - 1$

Hence $\frac{10}{\left(x - 1\right) \left({x}^{2} + 9\right)} = \frac{1}{x - 1} - \frac{x + 1}{{x}^{2} + 9}$