Q. 36 - Precalculus Enhanced with Graphing Utilities

Question

In Problem, Write the partial fraction decomposition of each rational expression.
$\frac{x}{\left(x^2+9\right)(x+1)}$

Step-by-Step Solution

Answer

Partial fraction decomposition of a rational expression $\frac{x}{\left(x^2+9\right)(x+1)}$is

$\frac{x}{\left(x^2+9\right)(x+1)}=\frac{\frac{-1}{10}}{(x+1)}+\frac{\frac{1}{10}x+\frac{9}{10}}{\left(x^2+9\right)}$

Step 1. Given data

The given rational expression is 

$\frac{x}{\left(x^2+9\right)(x+1)}$

Step 2. Decomposition of rational expression

Partial fraction decomposition of rational expression  

$$\begin{gathered} \frac{P\left(x\right)}{Q\left(x\right)}=\frac{A_1}{x-a_1}+\frac{A_2}{x-a_2}+\cdots +\frac{A_n}{x-a_n} \\ \frac{x}{\left(x^2+9\right)(x+1)}=\frac{A}{(x+1)}+\frac{Bx+C}{\left(x^2+9\right)} \\ \frac{x}{\left(x^2+9\right)(x+1)}=\frac{A\left(x^2+9\right)}{\left(x^2+9\right)(x+1)}+\frac{\left(Bx+C\right)(x+1)}{\left(x^2+9\right)(x+1)} \end{gathered}$$

Step 3. Value of numerator of partial fractions

Rearrange the equation

$$\begin{gathered} \frac{x}{\left(x^2+9\right)(x+1)}=\frac{A\left(x^2+9\right)}{\left(x^2+9\right)(x+1)}+\frac{\left(Bx+C\right)(x+1)}{\left(x^2+9\right)(x+1)} \\ x=A\left(x^2+9\right)+\left(Bx+C\right)(x+1) \\ 0x^2+1x+0=(A+B)x^2+(B+C)x+(9A+C) \end{gathered}$$

Equate the constants

$$\begin{gathered} 9A+C=0 \\ C=-9A \end{gathered}$$

Equate the coefficients of x

$$\begin{gathered} B+C=1 \\ B-9A=1 \\ B=1+9A \end{gathered}$$

Equate the coefficients of $x^2$

$$\begin{gathered} A+B=0 \\ A+1+9A=0 \\ 10A=-1 \\ A=\frac{-1}{10} \\ So B=\frac{1}{10}, C=\frac{9}{10} \end{gathered}$$

So partial fraction decomposition is  $\frac{x}{\left(x^2+9\right)(x+1)}=\frac{\frac{-1}{10}}{(x+1)}+\frac{\frac{1}{10}x+\frac{9}{10}}{\left(x^2+9\right)}$

Step 4. Verification

Simplify the partial fractions

$$\begin{gathered} \frac{\frac{-1}{10}}{(x+1)}+\frac{\frac{1}{10}x+\frac{9}{10}}{\left(x^2+9\right)}=\frac{\frac{-1}{10}\left(x^2+9\right)+\left(\frac{1}{10}x+\frac{9}{10}\right)(x+1)}{\left(x^2+9\right)(x+1)} \\ =\frac{x}{\left(x^2+9\right)(x+1)} \end{gathered}$$

It is equal to the original expression so partial fraction decomposition is correct.