Answer
\[V = 6\pi + 2\pi \ln \,\left( {\frac{2}{5}} \right)\]
Work Step by Step
\[\begin{gathered}
We\,\,have\,\,to\,\,find\,\,the\,\,volume\,\,of\,\,the\,\,solid\,\,obtained\,\,by \hfill \\
revolving\,\,the\,\,region\,\,bounded\,\,by \hfill \\
\hfill \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y = \frac{1}{{x + 2}}\,\,\,,\,\,\,\,y = 0\,\,\,\,,\,\,x = 0\,\,and\,\,x = 3 \hfill \\
\hfill \\
is\,\,revolved\,\,about\,\,x = - 1 \hfill \\
\hfill \\
V = 2\pi \int_0^3 {\,\left( {x + 1} \right)ydx} \hfill \\
\hfill \\
substitute\,\,\,\,\,\,\,\,y = \frac{1}{{x + 2}} \hfill \\
\hfill \\
V = 2\pi \int_0^3 {\,\left( {\frac{{x + 1}}{{x + 2}}} \right)dx} \hfill \\
\hfill \\
by\,\,the\,\,long\,\,division\,\,\frac{{x + 1}}{{x + 2}} = 1 - \frac{1}{{x + 2}} \hfill \\
\hfill \\
V = 2\pi \int_0^3 {\,\left( {1 - \frac{1}{{x + 2}}} \right)dx} \hfill \\
\hfill \\
integrate\,\,and\,\,evaluate \hfill \\
\hfill \\
V = 2\pi \,\,\left[ {x - \ln \,\left( {x + 2} \right)} \right]_0^3 \hfill \\
\hfill \\
V = 2\pi \,\,\left[ {3 - \ln 5 + \ln 2} \right] \hfill \\
\hfill \\
solution \hfill \\
\hfill \\
V = 6\pi + 2\pi \ln \,\left( {\frac{2}{5}} \right) \hfill \\
\end{gathered} \]