Chapter 11.5, Problem 33AYU

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

To determine

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

Answer to Problem 33AYU

Solution:

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

Explanation of Solution

Given:

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

Formula used:

If $Q\left(x\right) = \left(x - a_1\right)\dots \dots .\left(x \text{-} a_n\right)$ 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.

Calculation:

The denominator $x^2 + 2x - 3$ can be rewritten as $\left(x + 3\right)\left(x - 1\right)$.

The given partial fraction decomposition takes the form

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

$= >x = A\left(x - 1\right) + B\left(x + 3\right)$   ----- Eq (1)

Equating the co-efficients on both sides in Eq(1),

$\begin{array}{ccc}x, & & = >A + B = 1\end{array}$   -----Eq(2)

$constant = > \text{-} A + 3B = 0$   ----- Eq(3)

Solving Eq(2), Eq(3) for $A, B$ we get

$\text{A} = 3/4, \text{B} = 1/4$

From Eq (1), the partial fraction decomposition of

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