Chapter 7.3, Problem 53E

To determine

To evaluate the rational expression in its partial fraction decomposition form

Answer to Problem 53E

The partial decomposition form of the rational expression

  $\frac{7}{x^2-14x}$ is $\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{x}+\frac{\left(-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}{(x+14)}$

Explanation of Solution

Given information:

The rational expression is $\frac{7}{x^2-14x}$

Calculation:

On simplifying the expression,

  $\frac{7}{x^2-14x}=\frac{7}{x(x+14)}$

The form of the partial fraction decomposition can be written as,

  $\begin{array}{l}\frac{7}{x(x+14)}=\frac{A}{x}+\frac{B}{(x+14)}\\ \frac{7}{x(x+14)}=\frac{xB+(x+14)A}{x(x+14)}\end{array}$ (After cross multiplication)

Since the denominators are equal on both sides so the numerators on both sides can also be equated

Therefore,

  $\begin{array}{l}7=xB+(x+14)A\\ 7=xA+xB+14A\\ 7=x(A+B)+14A\end{array}$

Equating x term and constant term,

  $\begin{array}{l}A+B=0(1)\\ 14A=7(2)\end{array}$

  $\begin{array}{l}A=\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$14$}\right.\\ A=\frac{1}{2}\end{array}$ [From equation $(2)$ ]

  $\begin{array}{l}B=-A\\ B=-\frac{1}{2}\end{array}$ [From equation $(1)$ ]

Hence, the required partial fraction decomposition form is $\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{x}+\frac{\left(-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}{(x+14)}$