Chapter 9, Problem 7CLT

To determine

The partial fraction for the given rational expression.

Answer to Problem 7CLT

  $\frac{2x^2-x-6}{x^3+4x}=-\frac{3}{2x}+\frac{7x-2}{2(x^2+4)}$

Explanation of Solution

Given information: Given ration expression is

Formula used: To write the partial fraction we use this formula:

  $\frac{a}{x}+\frac{bx+c}{(x^2+4)}$

Calculation: Firstly we will find the factors of the denominator

  $x^3+4x=x(x^2+4)$

Now rational expression becomes

  $\frac{2x^2-x-6}{x^3+4x}=\frac{2x^2-x-6}{x(x^2+4)}$

Use the partial fraction formula

  $\begin{array}{l}\frac{2x^2-x-6}{x(x^2+4)}=\frac{a}{x}+\frac{bx+c}{(x^2+4)}\\ \text{ =}\frac{a(x^2+4)+x(bx+c)}{x(x^2+4)}\\ \text{ =}\frac{a{x}^2+4a+b{x}^2+cx}{x(x^2+4)}\\ \text{ =}\frac{(a+b)x^2+cx+4a}{x(x^2+4)}\end{array}$

Cancel the denominator both the sides

  $2x^2-x-6=(a+b)x^2+cx+4a$

Compare both the sides

  $\begin{array}{l}a+b=2---(1)\\ c=-1\\ 4a=-6\Rightarrow a=\frac{-3}{2}\\ Fromequation(1)\\ \frac{-3}{2}+b=2\\ b=\frac{7}{2}\end{array}$

Put the value of a, b & c

  $\begin{array}{l}\frac{2x^2-x-6}{x(x^2+4)}=\frac{a}{x}+\frac{bx+c}{(x^2+4)}\\ \frac{2x^2-x-6}{x(x^2+4)}=\frac{-3}{2x}+\frac{\frac{7}{2}x-1}{(x^2+4)}\\ \frac{2x^2-x-6}{x(x^2+4)}=\frac{-3}{2x}+\frac{7x-2}{2(x^2+4)}\end{array}$