Q. 37 - Precalculus Enhanced with Graphing Utilities

Question

In Problem, Write the partial fraction decomposition of each rational expression.
$\frac{x^3}{{\left(x^2+4\right)}^2}$

Step-by-Step Solution

Answer

Partial fraction decomposition of a rational expression $\frac{x^3}{{\left(x^2+4\right)}^2}$is

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

Step 1. Given data

The given rational expression is 

$\frac{x^3}{{\left(x^2+4\right)}^2}$

Step 2. Decomposition of rational expression

Partial fraction decomposition of rational expression  

$$\begin{gathered} \frac{P\left(x\right)}{Q\left(x\right)}=\frac{A_1}{x-a_1}+\frac{A_2}{x-a_2}+\cdots +\frac{A_n}{x-a_n} \\ \frac{x^3}{{\left(x^2+4\right)}^2}=\frac{Ax+B}{\left(x^2+4\right)}+\frac{Cx+D}{{\left(x^2+4\right)}^2} \\ \frac{x^3}{{\left(x^2+4\right)}^2}=\frac{\left(Ax+B\right)\left(x^2+4\right)}{\left(x^2+4\right)}+\frac{\left(Cx+D\right)}{{\left(x^2+4\right)}^2} \end{gathered}$$

Step 3. Value of numerator of partial fractions

Rearrange the equation

$$\begin{gathered} \frac{x^3}{{\left(x^2+4\right)}^2}=\frac{\left(Ax+B\right)\left(x^2+4\right)}{\left(x^2+4\right)}+\frac{\left(Cx+D\right)}{{\left(x^2+4\right)}^2} \\ {x}^3=\left(Ax+B\right)\left(x^2+4\right)+\left(Cx+D\right) \\ {x}^3=A{x}^3+4Ax+B{x}^2+4B+Cx+D \\ {x}^3+0x^2+0x+0=A{x}^3+B{x}^2+(4A+C)x+(4B+D) \end{gathered}$$

Equate the coefficients and constant

$$\begin{gathered} A=1 \\ B=0 \\ and 4A+C=0 \\ 4\left(1\right)+C=0 \\ C=-4 \\ and 4B+D=0 \\ 4\left(0\right)+D=0 \\ D=0 \end{gathered}$$

So partial fraction decomposition is $\frac{x^3}{{\left(x^2+4\right)}^2}=\frac{x}{\left(x^2+4\right)}+\frac{-4x}{{\left(x^2+4\right)}^2}$

Step 4. Verification

Simplify the partial fractions

$$\begin{gathered} \frac{x}{\left(x^2+4\right)}+\frac{-4x}{{\left(x^2+4\right)}^2}=\frac{x\left(x^2+4\right)-4x}{{\left(x^2+4\right)}^2} \\ =\frac{x^3}{{\left(x^2+4\right)}^2} \end{gathered}$$

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