Chapter 7, Problem 8CT

To determine

To write: the partial fraction decomposition for the rational function $\frac{x^3+x^2+x+2}{x^4-x^2}$ .

Answer to Problem 8CT

The partial fraction decomposition is given by.

  $\frac{x^3+x^2+x+1}{x^4-x^2}=-\frac{1}{x}-\frac{2}{x^2}-\frac{1}{2\left(x+1\right)}+\frac{5}{2\left(x-1\right)}$

Explanation of Solution

Given information :

Rational function is given as, $\frac{x^3+x^2+x+2}{x^4-x^2}$ .

Rational function is given as, $\frac{x^3+x^2+x+2}{x^4-x^2}$ .

  $\frac{x^3+x^2+x+2}{x^4-x^2}$

  $\frac{x^3+x^2+x+2}{x^4-x^2}=\frac{x^3+x^2+x+2}{x^2\left(x-1\right)\left(x+1\right)}$

  $\frac{x^3+x^2+x+2}{x^2\left(x-1\right)\left(x+1\right)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}+\frac{D}{x-1}$

  $\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}+\frac{D}{x-1}=\frac{Ax\left(x-1\right)\left(x+1\right)+B\left(x-1\right)\left(x+1\right)+x^2\left(x-1\right)C+x^2\left(x+1\right)D}{x^2\left(x-1\right)\left(x+1\right)}$

  $\frac{x^3+x^2+x+2}{x^4-x^2}=\frac{Ax\left(x-1\right)\left(x+1\right)+B\left(x-1\right)\left(x+1\right)+x^2\left(x-1\right)C+x^2\left(x+1\right)D}{x^2\left(x-1\right)\left(x+1\right)}$

  $x^3+x^2+x+2=Ax\left(x-1\right)\left(x+1\right)+B\left(x-1\right)\left(x+1\right)+x^2\left(x-1\right)C+x^2\left(x+1\right)D$

  $x^3+x^2+x+2=A{x}^3-Ax+B{x}^2-B+C{x}^3-C{x}^2+D{x}^3+D{x}^2$

  $x^3+x^2+x+2=\left(A+C+D\right)x^3-Ax+\left(B+D-C\right)x^2-B$

  $\text{After comparing,}$

  $\begin{array}{l}A+C+D=1\\ B-C+D=1\\ A=-1\\ B=-2\end{array}$

  $\text{After solving above equation,}$

  $\begin{array}{l}A=-1\\ B=-2\\ C=-\frac{1}{2}\\ D=\frac{5}{2}\end{array}$

So,

The partial fraction decomposition is given by.

  $\begin{array}{c}\frac{x^3+x^2+x+1}{x^4-x^2}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}+\frac{D}{x-1}\\ =-\frac{1}{x}-\frac{2}{x^2}-\frac{1}{2\left(x+1\right)}+\frac{5}{2\left(x-1\right)}\end{array}$