Chapter 7, Problem 64RE

To determine

Find the partial fraction for the given rational expression.

Answer to Problem 64RE

The partial fraction form is:

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

Explanation of Solution

Given:

The rational expression

  $\frac{4x}{3{(x-1)}^2}$

Consider the rational expression

  $\frac{4x}{3{(x-1)}^2}$

Since the degree of denominator is greater than numerator, the expression is in proper form.

Take one partial fraction with constant numerator for linear factor and for each power of (x−1) in denominator.

  $\frac{4x}{3{(x-1)}^2}=\frac{A}{3}+\frac{B}{x-1}+\frac{C}{{(x-1)}^2}$

Simplify as follows

  $\frac{4x}{3{(x-1)}^2}=\frac{A{(x-1)}^2+3B(x-1)+3C}{3{(x-1)}^2}$

Thus,

  $4x=A{(x-1)}^2+3B(x-1)+3C$ ......(1)

Substitute $x=1$ in equation (1) to get

  $\begin{array}{l}4\left(1\right)=A{\left(1–1\right)}^2+3B\left(1–1\right)+3C\hfill \\ 4=3C\hfill \\ \frac{4}{3}=C\hfill \end{array}$

Substitute $x=2,C=\frac{4}{3}$ in equation (1)to get

  $\begin{array}{l}4\left(2\right)=A{\left(2–1\right)}^2+3B\left(2–1\right)+3\left(\frac{4}{3}\right)\hfill \\ 8=A+3B+4\hfill \end{array}$

  $4=A+3B$ .....(2)

Substitute $x=0,C=\frac{4}{3}$ in equation (1) to get

  $\begin{array}{l}4\left(0\right)=A{\left(0–1\right)}^2+3B\left(0–1\right)+3C\hfill \\ 0=A–3B+3\frac{4}{3}\hfill \\ 0=A–3B+4\hfill \end{array}$

  $–4=A–3B$ .....(3)

Add equations (2) and (3)

  $\begin{array}{l}0=2A\hfill \\ 0=A\hfill \end{array}$

Substitute $A=0$ in $4=A+3B$

  $\begin{array}{l}4=0+3B\hfill \\ 4=3B\hfill \\ \frac{4}{3}=B\hfill \end{array}$

Hence, the partial fraction is given by

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