Answer
\[V = \pi \,\,\left[ {\frac{{24}}{5} - 2\ln 5} \right]\]
Work Step by Step
\[\begin{gathered}
Using\,the\,disk\,method \hfill \\
\hfill \\
V = \pi \,\,\left[ {\int_0^4 {\frac{{{x^2}}}{{\,{{\left( {1 + x} \right)}^2}}}dx} } \right] \hfill \\
\hfill \\
\frac{{{x^2}}}{{\,{{\left( {1 + x} \right)}^2}}} = \frac{{{x^2}}}{{1 + 2x + {x^2}}} \hfill \\
\hfill \\
{\text{using}}\,\,{\text{the}}\,\,\,{\text{long}}\,\,{\text{division}} \hfill \\
\frac{{{x^2}}}{{1 + 2x + {x^2}}} = 1 - \frac{{2x + 1}}{{{x^2} + 2x + 1}} \hfill \\
\hfill \\
then \hfill \\
\hfill \\
V = \pi \int_0^4 {\,\left( {1 - \frac{{2x + 1}}{{{x^2} + 2x + 1}}} \right)dx} \hfill \\
\hfill \\
or \hfill \\
\hfill \\
V = \int_0^4 {\,\left( {1 - \frac{{2x + 2 - 1}}{{{x^2} + 2x + 1}}} \right)} dx \hfill \\
\hfill \\
V = \pi \int_0^4 {\,\left( {1 - \frac{{2x + 2}}{{{x^2} + 2x + 1}} + \frac{1}{{\,{{\left( {x + 1} \right)}^2}}}} \right)} dx \hfill \\
\hfill \\
integrate\,\,and\,\,evaluate \hfill \\
\hfill \\
V = \pi \,\,\left[ {x - \ln \,{{\left( {x + 1} \right)}^2} - \frac{1}{{x + 1}}} \right]_0^4 \hfill \\
\hfill \\
solution \hfill \\
\hfill \\
V = \pi \,\,\left[ {\frac{{24}}{5} - 2\ln 5} \right] \hfill \\
\end{gathered} \]
