Chapter 5.7, Problem 30E

To determine

To calculate: The value of given integral by using long division

Answer to Problem 30E

The value of the given integral is $\frac{r^2}{2}-4r-16\ln \left|r+4\right|+C$

Explanation of Solution

Given information:

The integral is ${\displaystyle \int \frac{r^2}{r+4}dr}$

Formula used:

  ${\displaystyle \int \frac{1}{x}dx=\ln \left|x\right|+C}$

  ${\displaystyle \int dx=x+C}$

Calculation:

Use long division to rewrite the original integrand

  ${\displaystyle \int r-\frac{4r}{r+4}}dr$

Separate the integral into two smaller integrals and factor the 4 out of the second term

  ${\displaystyle \int rdr-4{\displaystyle \int \frac{r}{r+4}dr}}$

To simplify integration of the second term, it is necessary to remove $r$ from the numerator

  $\frac{r}{r+4}=\frac{r+4}{r+4}-\frac{4}{r+4}=1-\frac{4}{r+4}$

Substitute the new term into the second integral. Then split it into two separate integrals and again factor the 4 out of the integral

  ${\displaystyle \int rdr-4}\left({\displaystyle \int 1dr-4{\displaystyle \int \frac{1}{r+4}dr}}\right)$

Integrate using $u=r+4$ to solve the third integral

  $\frac{r^2}{4}-4\left(r-4\ln \left|r+4\right|\right)$

Distribute and add the constant $C$

  $\frac{r^2}{2}-4r-16\ln \left|r+4\right|+C$

Therefore, the value of the given integral is $\frac{r^2}{2}-4r-16\ln \left|r+4\right|+C$