Chapter 7, Problem 13CT

To determine

To write: the partial decomposition of the given expression and check the result algebraically.

Answer to Problem 13CT

The partial fraction is given by. $\frac{-5}{x}+\frac{3}{\left(x+1\right)}+\frac{3}{\left(x-1\right)}$ .

Explanation of Solution

Given:

The given rational expression is

  $\frac{x^2+5}{x^3-x}$

Calculation:

Consider the rational expression

  $\frac{x^2+5}{x^3-x}$

Factorizing the denominator,

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

Write the decomposition form,

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

Multiplying throughout by $x^3-x$

  $x^2+5=A\left(x-1\right)\left(x+1\right)+Bx\left(x-1\right)+Cx\left(x+1\right)$

Put $x=0$ in the above equation, we get,

  $\begin{array}{l}-A=5\\ A=-5\end{array}$

Put $x=1$ in the above equation, we get

  $\begin{array}{l}2C=6\\ C=3\end{array}$

Put $x=-1$ in the above equation, we get

  $\begin{array}{l}2B=6\\ B=3\end{array}$

SO,

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

So, the expression in partial fraction is obtained as: $\frac{-5}{x}+\frac{3}{\left(x+1\right)}+\frac{3}{\left(x-1\right)}$ .