Chapter 11, Problem 25CT

To determine

To find: The expression of $\frac{4x^2-3}{x{\left(x^2+3\right)}^2}$ .

Answer to Problem 25CT

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

Explanation of Solution

Given information:

The given expression is $\frac{4x^2-3}{x{\left(x^2+3\right)}^2}=\frac{\left(-\frac{1}{3}\right)}{x}+\frac{\left(\frac{1}{3}\right)x}{\left(x^2+3\right)}+\frac{5x}{{\left(x^2+3\right)}^2}$ .

Calculation:

Calculate the expression,

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

Equate the coefficients to form a new system of equation.

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

The fifth equation in the system gives the value of $A$ as $-\frac{3}{9}$ or $-\frac{1}{3}$

  $\begin{array}{c}-\frac{1}{3}+B=0\\ B=\frac{1}{3}\end{array}$

Calculate the $D$ , substitute $-\frac{1}{3}$ for $A$ and $\frac{1}{3}$ for $B$ in $6A+3B+D=4$ and simplify.

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

Substitute 0 for $C$ in $3C+E=0$ .

  $\begin{array}{c}3\left(0\right)+E=0\\ E=0\end{array}$

Replace the variables with their values in the partial friction decomposition.

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

Therefore, the partial friction decomposition is $\frac{\left(-\frac{1}{3}\right)}{x}+\frac{\left(\frac{1}{3}\right)x}{\left(x^2+3\right)}+\frac{5x}{{\left(x^2+3\right)}^2}$ .