# How do you write the partial fraction decomposition of the rational expression  (9x^2 + 1)/(x^2(x − 2)^2)?

Aug 11, 2018

$\frac{9 {x}^{2} + 1}{{x}^{2} {\left(x - 2\right)}^{2}} = \frac{1}{4 x} + \frac{1}{4 {x}^{2}} - \frac{1}{4 \left(x - 2\right)} + \frac{37}{4 {\left(x - 2\right)}^{2}}$

#### Explanation:

Given that we have squared linear factors in the denominator, the decomposition will take a form like:

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

Multiplying both sides by ${x}^{2} {\left(x - 2\right)}^{2}$ this becomes:

$9 {x}^{2} + 1 = A x {\left(x - 2\right)}^{2} + B {\left(x - 2\right)}^{2} + C {x}^{2} \left(x - 2\right) + D {x}^{2}$

Putting $x = 0$ we get:

$1 = 4 B \text{ }$ so $\text{ } B = \frac{1}{4}$

Putting $x = 2$ we get:

$37 = 4 D \text{ }$ so $\text{ } D = \frac{37}{4}$

Looking at the coefficient of $x$, we have:

$0 = 4 A - 4 B = 4 A - 1 \text{ }$ and hence $A = \frac{1}{4}$

Looking at the coefficient of ${x}^{3}$, we have:

$0 = A + C \text{ }$ and hence $C = - A = - \frac{1}{4}$

So:

$\frac{9 {x}^{2} + 1}{{x}^{2} {\left(x - 2\right)}^{2}} = \frac{1}{4 x} + \frac{1}{4 {x}^{2}} - \frac{1}{4 \left(x - 2\right)} + \frac{37}{4 {\left(x - 2\right)}^{2}}$