Chapter 7.4, Problem 47E

To determine

To calculate:

Write the partial fraction decomposition of the improper rational expression.

  $\frac{x^4}{{\left(x-1\right)}^3}$

Answer to Problem 47E

The partial fraction decomposition of the improper rational expression is $\frac{x^4}{{\left(x-1\right)}^3}=x+3+\frac{6}{x-1}+\frac{4}{{\left(x-1\right)}^2}+\frac{1}{{\left(x-1\right)}^3}$

Explanation of Solution

Given information:

  $\frac{x^4}{{\left(x-1\right)}^3}$ is the given equation.

Calculation:

Since the degree of the numerator is not less than degree of the denominator, perform polynomial long division.

  $\frac{x^4}{{\left(x-1\right)}^3}=x+3+\frac{6x^2-8x+3}{{\left(x-1\right)}^3}$

Now deal with $\frac{6x^2-8x+3}{{\left(x-1\right)}^3}$

The form of the partial fraction decomposition is

  $\frac{6x^2-8x+3}{{\left(x-1\right)}^3}=\frac{A}{x-1}+\frac{B}{{\left(x-1\right)}^2}+\frac{C}{{\left(x-1\right)}^3}$

Write the right hand side as a single fraction.

  $\frac{6x^2-8x+3}{{\left(x-1\right)}^3}=\frac{{\left(x-1\right)}^{2}A+\left(x-1\right)B+C}{{\left(x-1\right)}^3}$

The denominators are equal, so equate the numerators.

  $\begin{array}{l}6x^2-8x+3={\left(x-1\right)}^{2}A+\left(x-1\right)B+C\\ 6x^2-8x+3=x^{2}A-2xA+xB+A-B+C\\ 6x^2-8x+3=x^{2}A+x\left(-2A+B\right)+A-B+C\end{array}$

The coefficients are to be equated,

  $\begin{array}{l}A=6\\ -2A+B=-8\\ A-B+C=3\end{array}$

By solving them we get $A=6,B=4,C=1$

Therefore,

  $\begin{array}{l}\frac{6x^2-8x+3}{{\left(x-1\right)}^3}=\frac{6}{x-1}+\frac{4}{{\left(x-1\right)}^2}+\frac{1}{{\left(x-1\right)}^3}\\ \frac{x^4}{{\left(x-1\right)}^3}=x+3+\frac{6}{x-1}+\frac{4}{{\left(x-1\right)}^2}+\frac{1}{{\left(x-1\right)}^3}\end{array}$

Conclusion:

The partial fraction decomposition of the improper rational expression is $\frac{x^4}{{\left(x-1\right)}^3}=x+3+\frac{6}{x-1}+\frac{4}{{\left(x-1\right)}^2}+\frac{1}{{\left(x-1\right)}^3}$