Chapter 5.3, Problem 23E

To determine

To evaluate : the integral

Answer to Problem 23E

The value of ${\displaystyle \underset{1}{\overset{2}{\int }}\frac{v^3+3v^6}{v^4}}dv$ is $\ln 2+7$

Explanation of Solution

Given information : The integral is ${\displaystyle \underset{1}{\overset{2}{\int }}\frac{v^3+3v^6}{v^4}}dv$

Calculation :

  $\begin{array}{l}{\displaystyle \underset{1}{\overset{2}{\int }}\frac{v^3+3v^6}{v^4}}dv\\ \Rightarrow {\displaystyle \underset{1}{\overset{2}{\int }}\frac{v^3}{v^4}+\frac{3v^6}{v^4}}dv\\ \Rightarrow {\displaystyle \underset{1}{\overset{2}{\int }}\frac{1}{v}+3v^2}dv\\ \frac{1}{v}dv=\ln \left|v\right|+C\&{\displaystyle \int v^{n}dv=\frac{v^{n+1}}{n+1}}+C\\ {\displaystyle \underset{1}{\overset{2}{\int }}\frac{1}{v}+3v^2}dv={\left[\ln v+3\left(\frac{v^3}{3}\right)\right]}_1^2\\ {\left[\ln v+v^3\right]}_1^2=\left(\ln 2+2^3\right)-\left(\ln 1+1^3\right)\\ \Rightarrow \ln 2+8-1\\ \Rightarrow \ln 2+7\end{array}$