Chapter 7.5, Problem 47E

a.

To determine

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

Answer to Problem 47E

The integrals of ${\displaystyle \int \frac{5x}{{\left(x+3\right)}^2}dx}=5\ln \left(x+3\right)+\frac{15}{\left(x+3\right)}+c$

Explanation of Solution

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

Formula used: If

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

is a rational function with the degree of Pless 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 ${\displaystyle \int \frac{5x}{{\left(x+3\right)}^3}dx}$

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

the partial fraction decompositionof f is

  $\frac{5x}{{\left(x+3\right)}^3}=\frac{A_1}{x+3}+\frac{A_2}{{\left(x+3\right)}^2}+\frac{A_3}{{\left(x+3\right)}^3}$

And know that $A_1{(x+3)}^2+A_2\left(x+3\right)+A_3=5x$

Setting $x=-3$

  $A_3=-15$

Setting $x=0$

  $\begin{array}{l}9A_1+3A_2+A_3=0\\ A_3=-15\\ 9A_1+3A_2=15\\ 3A_1+A_2=5\end{array}$

Setting $x=-1$

  $\begin{array}{l}4A_1+2A_2+A_3=-5\\ A_3=-15\\ 4A_1+2A_2=10\\ 2A_1+A_2=5\end{array}$

Solve $3A_1+A_2=5$ and $2A_1+A_2=5$

  $\begin{array}{l}{A}_1=0\\ A_2=5\end{array}$

Thus

  $\begin{array}{l}\frac{5x}{{\left(x+3\right)}^2}=\frac{5}{x+3}-\frac{15}{{\left(x+3\right)}^2}\\ {\displaystyle \int \frac{5x}{{\left(x+3\right)}^2}dx}={\displaystyle \int \left(\frac{5}{x+3}-\frac{15}{{\left(x+3\right)}^2}\right)dx}\\ {\displaystyle \int \frac{5x}{{\left(x+3\right)}^2}dx}={\displaystyle \int \left(\frac{5}{x+3}-\frac{15}{{\left(x+3\right)}^2}\right)dx}\\ {\displaystyle \int \frac{5}{x+3}dx-{\displaystyle \int \frac{15}{{\left(x+3\right)}^2}dx=5\ln \left(x+3\right)+\frac{15}{\left(x+3\right)}+c}}\\ {\displaystyle \int \frac{5x}{{\left(x+3\right)}^2}dx}=5\ln \left(x+3\right)+\frac{15}{\left(x+3\right)}+c\end{array}$

b.

To determine

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

Answer to Problem 47E

The integrals of ${\displaystyle \int \frac{5x}{{\left(x+3\right)}^3}dx}=\frac{-5}{\left(x+3\right)}+\frac{15}{2}\frac{1}{{\left(x+3\right)}^2}+c$

Explanation of Solution

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

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

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

the partial fraction decompositionof f is

  $\frac{5x}{{\left(x+3\right)}^2}=\frac{A_1}{x+3}+\frac{A_2}{{\left(x+3\right)}^2}$

And $A_1(x+3)+A_2=5x$

Setting $x=-3$

  $A_2=-15$

Setting $x=0$

  $\begin{array}{l}3A_1+A_2=0\\ 3A_1-15=0\\ A_1=5\end{array}$

Thus

  $\begin{array}{l}\frac{5x}{{\left(x+3\right)}^3}=\frac{5}{{\left(x+3\right)}^2}-\frac{15}{{\left(x+3\right)}^3}\\ {\displaystyle \int \frac{5x}{{\left(x+3\right)}^3}dx}={\displaystyle \int \left(\frac{5}{{\left(x+3\right)}^2}-\frac{15}{{\left(x+3\right)}^3}\right)dx}\\ {\displaystyle \int \frac{5x}{{\left(x+3\right)}^3}dx}={\displaystyle \int \left(\frac{5}{{\left(x+3\right)}^2}-\frac{15}{{\left(x+3\right)}^3}\right)dx}\\ {\displaystyle \int \frac{5}{{\left(x+3\right)}^2}dx-{\displaystyle \int \frac{15}{{\left(x+3\right)}^3}dx=\frac{-5}{\left(x+3\right)}+\frac{15}{2}\frac{1}{{\left(x+3\right)}^2}+c}}\\ {\displaystyle \int \frac{5x}{{\left(x+3\right)}^3}dx}=\frac{-5}{\left(x+3\right)}+\frac{15}{2}\frac{1}{{\left(x+3\right)}^2}+c\end{array}$