Chapter 11.5, Problem 45AYU

In problems 13-46, find the partial fraction decomposition of each rational expression. $\frac{2x+3}{x^4-9x^2}$

To determine

To find: The partial fraction decomposition of the given rational function.

Answer to Problem 45AYU

Solution:

$- \frac{2}{9x} - \frac{1}{3x^2} + \frac{1}{18\left(x + 3\right)} + \frac{1}{6\left(x - 3\right)}$

Explanation of Solution

Given:

$\frac{2x + 3}{x^4 - 9x^2}$

Formula used:

a. If $Q(x) = (x - a_1)\dots \dots .(x - a_n)$ then partial fraction decomposition of $\text{P}\left(\text{x}\right)/\text{Q}\left(\text{x}\right)$ takes the form:

$\frac{P\left(x\right)}{Q\left(x\right)} = \frac{A_1}{\left(x - a_1\right)} + \frac{A_2}{\left(x - a_2\right)} + \cdots \frac{A_n}{\left(x - a_n\right)}$

where, $A_1, A_2\dots A_n$ are to be determined.

b. If $Q$ has repeated linear factor of the form ${\left(x - a\right)}^n, \text{n} \ge 2$ an integer, then, in the partial fraction decomposition of $P(x)/Q(x)$, takes the form:

$\frac{P\left(x\right)}{Q\left(x\right)} = \frac{A_1}{\left(x - a\right)} + \frac{A_2}{{\left(x - a\right)}^2} + \cdots + \frac{A_n}{{\left(x - a\right)}^n}$

where, the numbers $A_1, A_2\dots A_n$ are to be determined.

Calculation:

The denominator $x^4 - 9x^2$ can be factored as $x^2\left(x + 3\right)\left(x - 3\right)$

The denominator has one repeated linear irreducible factor and two no repeated linear factor. Therefore the partial fraction decomposition takes the form:

$\frac{2x + 3}{x^4 - 9x^2} = \frac{2x + 3}{x^2\left(x + 3\right)\left(x - 3\right)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{\left(x + 3\right)} + \frac{D}{\left(x - 3\right)}$

$\Rightarrow 2x + 3 = Ax\left(x + 3\right)\left(x - 3\right) + B\left(x + 3\right)\left(x - 3\right) + C{x}^2\left(x - 3\right) + D{x}^2\left(x + 3\right)$   -----Eq (1)

Substituting $x = 0$, in Eq(1) we get $\text{B} = - 1/3$

Substituting $x = 3$, in Eq(1) we get $\text{D} = 1/6$

Substituting $x = - 3$, in Eq(1) we get $\text{C} = 1/18$

Equating coefficient of ${\text{x}}^3$ $\text{A} + \text{C} + \text{D} = 0$ or $\text{A} = - 2/9$

The partial fraction decomposition of $\frac{2x + 3}{x^4 - 9x^2}$ is $- \frac{2}{9x} - \frac{1}{3x^2} + \frac{1}{18\left(x + 3\right)} + \frac{1}{6\left(x - 3\right)}$