Chapter 11.5, Problem 23AYU

In problems 13-46, find the partial fraction decomposition of each rational expression. $\frac{x^2}{{\left(x-1\right)}^2{\left(x+1\right)}^2}$

To determine

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

Answer to Problem 23AYU

Solution:

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

Explanation of Solution

Given:

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

Formula used:

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 $P\left(x\right)\text{/}Q\left(x\right)$, 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 contains the repeated linear factors ${\left(x - 1\right)}^2 \& {\left(x + 1\right)}^2$

The given partial fraction decomposition takes the form

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

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

Let $x = 1$ in Eq(2)

$1 = 4\text{B}$ or $\text{B} = 1/4$

Let $x = - 1$ in Eq(2)

$D{\left( - 2\right)}^2 = 1$ or $\text{D} = 1/4$

$\begin{array}{lll}{x}^2\hfill & = \hfill & A\left(x - 1\right)\left(x^2 + 2x + 1\right) + B\left(x^2 + 2x + 1\right) + c\left(x^2 - 2x + 1\right)\left(x + 1\right) + D\left(x^2 - 2x + 1\right)\hfill \\ \hfill & = \hfill & A\left(x^3 + 2x^2 + x - x^2 - 2x - 1\right) + B\left(x^2 + 2x + 1\right) + C\left(x^3 - 2x^2 + x + x^2 - 2x + 1\right) + D\left(x^2 - 2x + 1\right)\hfill \end{array}$

Equating the co-efficients on both sides,

$x^3, \Rightarrow \text{A} + \text{C} = 0$   -----Eq(3)

$x^2, \Rightarrow \text{A} + \text{B} - \text{C} + \text{D} = 1$   -----Eq(4)

$x, \Rightarrow - \text{A} + 2\text{B} - \text{C} - 2\text{D} = 0$   -----Eq(5)

$\text{constant}\Rightarrow - \text{A} + \text{B} + \text{C} + \text{D} = 0$   -----Eq (6)

Solving for $\text{A}, \text{B}, \text{C}, \text{D}$ we get

$\text{A} = 1/4, \text{B} = 1/4, \text{C} = - 1/4 \text{and} \text{D} = 1/4$

From Eq (1), the partial fraction decomposition of

$\frac{x^2}{{\left(x - 1\right)}^2{\left(x + 1\right)}^2}$ is $\frac{1}{4\left(x - 1\right)} + \frac{1}{4{\left(x - 1\right)}^2} - \frac{1}{4\left(x + 1\right)} + \frac{1}{4{\left(x + 1\right)}^2}$.