Q. 25 - Precalculus Enhanced with Graphing Utilities

Question

In Problems $24$ and $25$, write the partial fraction decomposition of each rational expression.

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

Step-by-Step Solution

Answer

Partial fraction decomposition is $\frac{-\frac{1}{3}}{x}+\frac{\frac{1}{3}x}{\left(x^2+3\right)}+\frac{5x}{{\left(x^2+3\right)}^2}$

Step 1. Given information

An expression is given as $\frac{4x^2-3}{x{\left(x^2+3\right)}^2}$

Step 2. Decomposition fraction

First write the expression as

$\frac{4x^2-3}{x{\left(x^2+3\right)}^2}=\frac{A}{x}+\frac{Bx+C}{\left(x^2+3\right)}+\frac{Dx+E}{{\left(x^2+3\right)}^2}$

Now multiply $x{\left(x^2+3\right)}^2$ on both side,

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

Now simplify,

$$\begin{gathered} =A\left(x^4+6x^2+9\right)+(Bx+C)\left(x^3+3x\right)+(Dx+E)\left(x\right) \\ 4x^2-3=(A+B)x^4+C{x}^3+(6A+3B+D)x^2+(3C+E)x+9A \end{gathered}$$

Step 3. System of equations

Equate the coefficients to form new set of equations,

$\left\{\begin{array}{l}A+B=0\\ C=0\\ 6A+3B+D=4\\ 3C+E=0\\ 9A=-3\end{array}\right.$

From above set $A=-\frac{1}{3}$ and $B=\frac{1}{3}$

Substitute the values of A and B in the equation $6 A+3 B+D=4$

$$\begin{gathered} 6\left(-\frac{1}{3}\right)+3\left(\frac{1}{3}\right)+D=4 \\ -2+1+D=4 \\ -1+D=4 \\ D=5 \end{gathered}$$

Step 4. Solving set of equations

Now substitute the value in the equation

 $$\begin{gathered} 3 C+E=0 \\ 3\left(0\right)+E=0 \\ E=0 \end{gathered}$$

Now substitute the value in first equation,

$$\begin{gathered} \frac{A}{x}+\frac{Bx+C}{\left(x^2+3\right)}+\frac{Dx+E}{{\left(x^2+3\right)}^2}=\frac{\left(-\frac{1}{3}\right)}{x}+\frac{\left(\frac{1}{3}\right)x+\left(0\right)}{\left(x^2+3\right)}+\frac{5x+0}{{\left(x^2+3\right)}^2} \\ =\frac{\left(-\frac{1}{3}\right)}{x}+\frac{\left(\frac{1}{3}\right)x}{\left(x^2+3\right)}+\frac{5x}{{\left(x^2+3\right)}^2} \end{gathered}$$