Chapter 7.4, Problem 18E

To determine

To write the partial fraction decomposition of the rational expression.

Answer to Problem 18E

The partial fraction decomposition of the given fraction is,

  $\frac{-1}{x}+\frac{1}{\left(x-3\right)}$ .

Explanation of Solution

Given information:

The rational expression is $\frac{3}{x^2-3x}$ .

Concept used:

If fraction is $\frac{1}{\left(y+a\right)\left(y+b\right)}$ then, the partial fraction decomposition is $\frac{A}{\left(y+a\right)}+\frac{B}{\left(y+b\right)}$

Calculation:

  $\frac{3}{x^2-3x}$

Take x common from denominator.

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

Partial fraction decomposition is,

  $\begin{array}{l}\frac{3}{x\left(x-3\right)}=\frac{A}{x}+\frac{B}{\left(x-3\right)}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\mathrm{...............}\left(1\right)\text{}\text{}\text{}\\ \frac{3}{x\left(x-3\right)}=\frac{A\left(x-3\right)+Bx}{x\left(x-3\right)}\end{array}$

Denominators are same.

So, compare the numerators

  $\begin{array}{l}3=A\left(x-3\right)+Bx\\ 3=Ax-3A+Bx\\ 3=x\left(A+B\right)-3A\end{array}$

Compare the coefficient of $x$ .

  $A+B=0$

And compare the constant terms.

  $\begin{array}{c}-3A=3\\ \frac{-3A}{-3}=\frac{3}{-3}\\ A=-1\end{array}$

Substitute $A=-1$ in $A+B=0$ .

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

Substitute $A=-1$ in $B=1$ in equation (1).

  $\begin{array}{l}\frac{3}{x\left(x-3\right)}=\frac{A}{x}+\frac{B}{\left(x-3\right)}\\ \frac{3}{x\left(x-3\right)}=\frac{-1}{x}+\frac{1}{\left(x-3\right)}\end{array}$

So, the partial fraction decomposition of the given fraction is,

  $\frac{-1}{x}+\frac{1}{\left(x-3\right)}$ .