## Calculus: Early Transcendentals (2nd Edition)

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