Chapter 7.3, Problem 66E

To determine

To write the Partial fraction decomposition of the rational expression $\frac{x^2+12x-9}{x^3-9x}$

Answer to Problem 66E

The partial fraction of expansion is $\frac{1}{x}+\frac{12}{x^2-9}$

Explanation of Solution

Given information:

The rational expression is $\frac{x^2+12x-9}{x^3-9x}$

The rational expression is $\frac{x^2+12x-9}{x^3-9x}$

Here the denominator can be factorised and the given rational expression can be expressed as sum of partial fraction.

  $\frac{x^2+12x-9}{x^3-9x}=\frac{x^2+12x-9}{x\left(x^2-9\right)}=\frac{A}{x}+\frac{Bx+C}{x^2-9}$

where $A$ , $B$ and $C$ are the coefficients.

Now , for finding the value of coefficients $A$ , $B$ and $C$

  $\begin{array}{l}\frac{x^2+12x-9}{x^3-9x}=\frac{A}{x}+\frac{Bx+C}{x^2-9}=\frac{A\left(x^2-9\right)+B{x}^2+Cx}{x\left(x^2-9\right)}=\frac{x^2\left(A+B\right)+Cx-9A}{x\left(x^2-9\right)}\\ x^2+12x-9=x^2\left(A+B\right)+Cx-9A\end{array}$

Comparing coefficients of $x$

  $C=12$

Comparing constant terms in the equation

  $\begin{array}{l}-9A=-9\\ A=1\end{array}$

Comparing coefficients of $x^2$

  $\begin{array}{l}A+B=1\\ B=1-A\\ B=0\end{array}$

therefore , $\frac{x^2+12x-9}{x^3-9x}=\frac{1}{x}+\frac{12}{x^2-9}$

Checking algebraically,

  $\frac{1}{x}+\frac{12}{x^2-9}=\frac{\left(x^2-9\right)+12x}{x\left(x^2-9\right)}=\frac{x^2+12x-9}{x\left(x^2-9\right)}=\frac{x^2+12x-9}{x^3-9x}$

Checking Graphically,

  $\frac{x^2+12x-9}{x^3-9x}$

  

  $\frac{1}{x}+\frac{12}{x^2-9}$

  

The graph of rational expression and partial fraction is same hence graphically it is proved that

  $\frac{x^2+12x-9}{x^3-9x}=\frac{1}{x}+\frac{12}{x^2-9}$