Chapter 7.3, Problem 71E

To determine

To Find : Partial fraction of rational expression $\frac{x^4}{{(x-1)}^3}$

Answer to Problem 71E

The partial fraction of expansion is $x+3+\frac{6}{(x-1)}+\frac{4}{{(x-1)}^2}+\frac{1}{{(x-1)}^3}$

Explanation of Solution

Given information:

The rational expression is $\frac{x^4}{{(x-1)}^3}$

The rational expression is $\frac{x^4}{{(x-1)}^3}$

Here $N\left(x\right)/D\left(x\right)$ is an improper fractionbecause degree of $N\left(x\right)$ $>$ degree of $D\left(x\right)$ so divide $N\left(x\right)$ by $D\left(x\right)$ to obtain

  $\frac{x^4}{{(x-1)}^3}=x+3+\frac{6x^2-8x+3}{{(x-1)}^3}=x+3+\frac{A}{(x-1)}+\frac{B}{{(x-1)}^2}+\frac{C}{{(x-1)}^3}$

Now express $\frac{6x^2-8x+3}{{(x-1)}^3}$ in terms of partial fraction

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

Now , for finding the value of coefficients $A$ and $B$

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

Comparing coefficients of $x^2$

  $A=6$

Comparing coefficients of $x$

  $\begin{array}{l}-2A+B=-8\\ B=4\end{array}$

Comparing constant terms in the equation

  $\begin{array}{l}A-B+C=3\\ C=1\end{array}$

therefore , $\frac{x^4}{{(x-1)}^3}=x+3+\frac{6}{(x-1)}+\frac{4}{{(x-1)}^2}+\frac{1}{{(x-1)}^3}$

Checking algebraically,

  $\begin{array}{l}x+3+\frac{6}{(x-1)}+\frac{4}{{(x-1)}^2}+\frac{1}{{(x-1)}^3}=\frac{\left(x+3\right){\left(x-1\right)}^3+6{\left(x-1\right)}^2+4\left(x-1\right)+1}{{(x-1)}^3}\\ =\frac{x^4-3x^3+3x^2-x+3x^3-9x^2+9x-3+6x^2-12x+6+4x-4+1}{{(x-1)}^3}=\frac{x^4}{{(x-1)}^3}\end{array}$

Checking Graphically,

  $\frac{x^4}{{(x-1)}^3}$

  

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

  

Interpretation:

The graph of rational expression and partial fraction is same hence graphically it is proved that $\frac{x^4}{{(x-1)}^3}=x+3+\frac{6}{(x-1)}+\frac{4}{{(x-1)}^2}+\frac{1}{{(x-1)}^3}$