Chapter 7.5, Problem 48E

a.

To determine

To prove: Show that $A{\left(x-1\right)}^2+B\left(x-1\right)+c=x^2+3x+5$

Explanation of Solution

Given information:

The partial fraction decomposition of $f(x)$ is

  $\frac{x^2+3x+5}{{\left(x-1\right)}^3}=\frac{A}{\left(x-1\right)}+\frac{B}{{\left(x-1\right)}^2}+\frac{C}{{\left(x-1\right)}^3}$

Formula used: If

  $f(x)=\frac{P(x)}{{\left(x-r\right)}^m}$

is a rational function with the degree of P less than $m$ , then the partial fraction decompositionof fis

  $f(x)=\frac{A_1}{x-r}+\frac{A_2}{{\left(x-r\right)}^2}+\mathrm{.....}+\frac{A_m}{{\left(x-r\right)}^m}$

Calculation:

Function

  $f(x)=\frac{x^2+3x+5}{{\left(x-1\right)}^3}$

the partial fraction decompositionof $f(x)$ is

  $\frac{x^2+3x+5}{{\left(x-1\right)}^3}=\frac{A}{\left(x-1\right)}+\frac{B}{{\left(x-1\right)}^2}+\frac{C}{{\left(x-1\right)}^3}$

Multiplyboth side of the equation ${\left(x-1\right)}^3$

  $\begin{array}{l}{x}^2+3x+5=A{\left(x-1\right)}^2+B\left(x-1\right)+c\\ A{\left(x-1\right)}^2+B\left(x-1\right)+c=x^2+3x+5\end{array}$

b.

To determine

To Find: The value of $A,B,\text{and}C$

Answer to Problem 48E

The value of $A,B,\text{and}C$ are

  $\begin{array}{l}A=1\\ B=5\\ C=9\end{array}$

Explanation of Solution

Given information:

the partial fraction decompositionof $f(x)$ is

  $\frac{x^2+3x+5}{{\left(x-1\right)}^3}=\frac{A}{\left(x-1\right)}+\frac{B}{{\left(x-1\right)}^2}+\frac{C}{{\left(x-1\right)}^3}$

And $A{\left(x-1\right)}^2+B\left(x-1\right)+c=x^2+3x+5$

Formula used: If

  $f(x)=\frac{P(x)}{{\left(x-r\right)}^m}$

is a rational function with the degree of P less than $m$ , then the partial fraction decompositionof fis

  $f(x)=\frac{A_1}{x-r}+\frac{A_2}{{\left(x-r\right)}^2}+\mathrm{.....}+\frac{A_m}{{\left(x-r\right)}^m}$

Calculation:

Given

  $A{\left(x-1\right)}^2+B\left(x-1\right)+c=x^2+3x+5$

Expand and equate coefficients of like terms

  $\begin{array}{l}A{x}^2+x\left(B-2A\right)+A-B+C=x^2+3x+5\\ A=1\\ B-2A=3\\ A-B+C=5\end{array}$

Put $A=1$ in $B-2A=3$

  $B=5$

Similarly, $C=9$

c.

To determine

To Find:The integrals of ${\displaystyle \int \frac{x^2+3x+5}{{\left(x-1\right)}^3}dx}$

Answer to Problem 48E

The integrals of ${\displaystyle \int \frac{x^2+3x+5}{{\left(x-1\right)}^3}dx}=\ln \left|x-1\right|-\frac{5}{\left(x-1\right)}-\frac{9}{2{\left(x-1\right)}^2}+c$

Explanation of Solution

Given information: Use partial fraction to find the integrals of ${\displaystyle \int \frac{x^2+3x+5}{{\left(x-1\right)}^3}dx}$

Formula used: If

  $f(x)=\frac{P(x)}{{\left(x-r\right)}^m}$

is a rational function with the degree of P less than $m$ , then the partial fraction decompositionof fis

  $f(x)=\frac{A_1}{x-r}+\frac{A_2}{{\left(x-r\right)}^2}+\mathrm{.....}+\frac{A_m}{{\left(x-r\right)}^m}$

Calculation:

The integrals of ${\displaystyle \int \frac{x^2+3x+5}{{\left(x-1\right)}^3}dx}$

  $\begin{array}{l}f(x)=\frac{x^2+3x+5}{{\left(x-1\right)}^3}\\ \frac{x^2+3x+5}{{\left(x-1\right)}^3}=\frac{A}{x-1}+\frac{B}{{\left(x-1\right)}^2}+\frac{C}{{\left(x-1\right)}^3}\end{array}$

  $\begin{array}{l}\text{Given}\\ A=1\\ B=5\\ C=9\end{array}$

  $\begin{array}{l}\frac{x^2+3x+5}{{\left(x-1\right)}^3}=\frac{1}{x-1}+\frac{5}{{\left(x-1\right)}^2}+\frac{9}{{\left(x-1\right)}^3}\\ {\displaystyle \int \frac{x^2+3x+5}{{\left(x-1\right)}^3}dx}={\displaystyle \int \left(\frac{1}{x-1}+\frac{5}{{\left(x-1\right)}^2}+\frac{9}{{\left(x-1\right)}^3}\right)dx}\end{array}$

  ${\displaystyle \int \left(\frac{1}{x-1}+\frac{5}{{\left(x-1\right)}^2}+\frac{9}{{\left(x-1\right)}^3}\right)dx}={\displaystyle \int \frac{1}{\left(x-1\right)}dx+{\displaystyle \int \frac{5}{{\left(x-1\right)}^2}}}dx+{\displaystyle \int \frac{9}{{\left(x-1\right)}^3}}$

  ${\displaystyle \int \frac{1}{\left(x-1\right)}dx+{\displaystyle \int \frac{5}{{\left(x-1\right)}^2}}}dx+{\displaystyle \int \frac{9}{{\left(x-1\right)}^3}}=\ln \left|x-1\right|-\frac{5}{\left(x-1\right)}-\frac{9}{2{\left(x-1\right)}^2}+c$

  ${\displaystyle \int \frac{x^2+3x+5}{{\left(x-1\right)}^3}dx}=\ln \left|x-1\right|-\frac{5}{\left(x-1\right)}-\frac{9}{2{\left(x-1\right)}^2}+c$