# How do you express 1 / ((x+5)^2 (x-1))  in partial fractions?

##### 1 Answer
Sep 30, 2016

$\frac{1}{{\left(x + 5\right)}^{2} \left(x - 1\right)} = - \frac{1}{6 {\left(x + 5\right)}^{2}} - \frac{1}{36 \left(x + 5\right)} + \frac{1}{36 \left(x - 1\right)}$

#### Explanation:

$\frac{1}{{\left(x + 5\right)}^{2} \left(x - 1\right)} = \frac{A}{x + 5} ^ 2 + \frac{B}{x + 5} + \frac{C}{x - 1}$

Multiplying both sides of this equation by ${\left(x + 5\right)}^{2}$ we get:

$\frac{1}{x - 1} = A + B \left(x + 5\right) + \frac{C {\left(x + 5\right)}^{2}}{x - 1}$

Now let $x = - 5$ to find:

$A = \frac{1}{\left(- 5\right) - 1} = - \frac{1}{6}$

Multiplying both sides of the first equation by $\left(x - 1\right)$ we get:

$\frac{1}{x + 5} ^ 2 = \frac{A \left(x - 1\right)}{x + 5} ^ 2 + \frac{B \left(x - 1\right)}{x + 5} + C$

Now let $x = 1$ to find:

$C = \frac{1}{\left(1\right) + 5} ^ 2 = \frac{1}{36}$

Going back to the first equation and making a common denominator we find:

$\frac{1}{{\left(x + 5\right)}^{2} \left(x - 1\right)} = \frac{A}{x + 5} ^ 2 + \frac{B}{x + 5} + \frac{C}{x - 1}$

$\textcolor{w h i t e}{\frac{1}{{\left(x + 5\right)}^{2} \left(x - 1\right)}} = \frac{A \left(x - 1\right) + B \left(x - 1\right) \left(x + 5\right) + C {\left(x + 5\right)}^{2}}{{\left(x + 5\right)}^{2} \left(x - 1\right)}$

So equating the coefficients of ${x}^{2}$ in the numerator, we find:

$B = - C = - \frac{1}{36}$