Chapter 11.5, Problem 20AYU

To determine

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

Answer to Problem 20AYU

Solution:

$\frac{3}{4\left(x - 2\right)} - \frac{3}{4x} - \frac{1}{2x^2}$

Explanation of Solution

Given:

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

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 $P\left(x\right)\text{/}Q\left(x\right)$, 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 given partial fraction decomposition takes the form

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

$\begin{array}{l}x + 1 = A{x}^2 + B\left(x - 2\right)x + C\left(x - 2\right)\\ = A{x}^2 - 2Bx + B{x}^2 + Cx - 2C\end{array}$

Equating the coefficients of $x^2$, x and constant we get

$\begin{array}{rrr}\hfill A + B& \hfill = & \hfill 0\\ \hfill - 2B + C& \hfill = & \hfill 1\\ \hfill - 2C& \hfill = & \hfill 1\end{array}$

Solving for $A, B, C$

$\begin{array}{rrr}\hfill A& \hfill = & \hfill 3/4\\ \hfill B& \hfill = & \hfill - 3/4\\ \hfill C& \hfill = & \hfill - 1/2\end{array}$

From Eq (1), the partial fraction decomposition of

$\frac{x + 1}{x^2\left(x - 2\right)}$ is $\frac{3}{4\left(x - 2\right)} - \frac{3}{4x} - \frac{1}{2x^2}$.