Chapter 11.5, Problem 40AYU

In problems 13-46, find the partial fraction decomposition of each rational expression. $\frac{x^2+1}{x^3+x^2-5x+3}$

To determine

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

Answer to Problem 40AYU

Solution:

$\frac{3}{8 \left(x - 1\right)} + \frac{1}{2{\left(x - 1\right)}^2} + \frac{5}{8 \left(x + 3\right)}$

Explanation of Solution

Given:

$\frac{x^2 + 1}{x^3 + x^2 - 5x + 3}$

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 $P\left(x\right)/Q\left(x\right)$, 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 of the rational function $x^3 + x^2 - 5x + 3$ can be factored as ${\left(x - 1\right)}^2\left(x + 3\right)$

The denominator has repeated linear factor $\left(\text{x−1}\right)$ and non repeated linear factor $\left(\text{x} + 3\right)$. Therefore the partial fraction decomposition takes the form

$\frac{x^2 + 1}{x^3 + x^2 - 5x + 3} = \frac{A}{x - 1} + \frac{B}{{\left(x - 1\right)}^2} + \frac{C}{\left(x + 3\right)}$

$\begin{array}{c}\Rightarrow x^2 + 1 = A\left(x - 1\right)\left(x + 3\right) + B\left(x + 3\right) + C{\left(x - 1\right)}^2\\ = A\left(x^2 + 3x - 3\right) + B\left(x + 3\right) + C\left(x^2 + 1 - 2x\right)\end{array}$   -----Eq (1)

Substituting $\text{x} = 1$ in Eq 1, we get $\text{B} = 1/2$

Substituting $\text{x} = - 3$ in Eq 1, we get $\text{C} = 5/8$ and

Substituting $\text{x} = 0$ in Eq 1, we get $- 3\text{A} + 3\text{B} + \text{C} = 1$ or $\text{A} = 3/8$

The partial fraction decomposition of $\frac{x^2 + 1}{x^3 + x^2 - 5x + 3}$ is $\frac{3}{8 \left(x - 1\right)} + \frac{1}{2{\left(x - 1\right)}^2} + \frac{5}{8 \left(x + 3\right)}$