Chapter 7.4, Problem 20E

To determine

To write:The partial fraction decomposition of the rational expression $\frac{x+1}{x^2-x-6}$ .

Answer to Problem 20E

  $\frac{x+1}{x^2-x-6}=\frac{1}{5\left(x+2\right)}+\frac{4}{5\left(x-3\right)}$

Explanation of Solution

Given:

  $\frac{x+1}{x^2-x-6}$

Calculation:

The given rational expression is

  $\frac{x+1}{x^2-x-6}$

The factor of the denominator is given by $x^2-x-6=\left(x-3\right)\left(x+2\right)$

So, the rational expression becomes $\frac{x+1}{\left(x+2\right)\left(x-3\right)}$

The form of partial fraction decomposition of this expression is given by

  $\frac{x+1}{x^2-x-6}=\frac{A}{x+2}+\frac{B}{x-3}$

On simplifying

  $x+1=A\left(x-3\right)+B\left(x+2\right)$

Now, compare the coefficient of x ,

  $\begin{array}{l}1=A+B\\ \Rightarrow A+B=\mathrm{1...}\left(i\right)\end{array}$

And comparing the constant term on both sides

  $1=-3A+2B\mathrm{...}\left(ii\right)$

Solving equations $\left(i\right)$ and $\left(ii\right)$

  $\begin{array}{c}1=-3A+2\left(1-A\right)\\ 1=-3A+2-2A\\ 5A=1\\ A=\frac{1}{5}\end{array}$

Therefore, using equation $\left(i\right)$

  $\begin{array}{c}\frac{1}{5}+B=1\\ B=1-\frac{1}{5}\\ B=\frac{4}{5}\end{array}$

Therefore, the partial fraction decomposition is given by

  $\begin{array}{l}\frac{x+1}{x^2-x-6}=\frac{\frac{1}{5}}{x+2}+\frac{\frac{4}{5}}{x-3}\\ \frac{x+1}{x^2-x-6}=\frac{1}{5\left(x+2\right)}+\frac{4}{5\left(x-3\right)}\end{array}$

In order to check the result algebraically, simplify the expression on right hand side of the equation

  $\begin{array}{c}\frac{1}{5\left(x+2\right)}+\frac{4}{5\left(x-3\right)}=\frac{x-3+4x+8}{5\left(x+2\right)\left(x-3\right)}\\ =\frac{5x+5}{5\left(x+2\right)\left(x-3\right)}\\ =\frac{x+1}{x^2-x-6}\end{array}$

This resultant expression is same as the given rational expression. Hence, the partial fraction of the given expression is true.