Chapter 7, Problem 62RE

To determine

Find the partial fraction for the given rational expression.

Answer to Problem 62RE

The partial fraction form is:

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

Explanation of Solution

Given:

The rational expression

  $\frac{9}{x^2-9}$

Consider the rational expression

  $\frac{9}{x^2-9}$

Since the degree of denominator is greater than numerator, the expression is in the proper form.

The denominator can be written as

  $x^2–9=\left(x–3\right)\left(x+3\right)$

Take one partial fraction with constant numerator for each linear factor of denominator.

  $\frac{9}{x^2-9}=\frac{9}{(x-3)(x+3)}$

  $=\frac{A}{x-3}+\frac{B}{x+3}$

Simplify as follows

  $\frac{9}{x^2-9}=\frac{A(x+3)+B(x-3)}{(x-3)(x+3)}$

Thus,

  $9=A\left(x+3\right)+B\left(x–3\right)$ .....(1)

Substitute $x=–3$ in equation (1) to get

  $\begin{array}{l}9=A\left(–3+3\right)+B\left(–3–3\right)\hfill \\ 9=B\left(–6\right)\hfill \\ -\frac{3}{2}=B\hfill \end{array}$

Substitute $x=3$ in equation (1) to get

  $\begin{array}{l}9=A\left(3+3\right)+B\left(3–3\right)\hfill \\ 9=6A\hfill \\ \frac{3}{2}=A\hfill \end{array}$

Hence, the partial fraction is given by

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