\(V = \pi \int\limits_0^1 {{x^2}\ln (x + 1)dx}\)

Đặt: \(\left\{ \begin{array}{l} u = \ln (x + 1)\\ dv = {x^2}dx \end{array} \right. \Rightarrow \left\{ \begin{array}{l} du = \frac{1}{{x + 1}}dx\\ v = \frac{1}{3}{x^3} \end{array} \right.\)

Vậy:

\(\begin{array}{*{20}{l}}
\begin{array}{l}
V = \pi \left[ {\left. {\frac{{{x^3}}}{3}\ln (x + 1)} \right|_0^1 - \frac{1}{3}\int\limits_0^1 {\frac{{{x^3}}}{{x + 1}}dx} } \right]\\
 = \pi \left[ {\frac{1}{3}\ln 2 - \int\limits_0^1 {\left( {\frac{{{x^3} + 1}}{{x + 1}} - \frac{1}{{x + 1}}} \right)dx} } \right]
\end{array}\\
\begin{array}{l}
 = \pi \left[ {\frac{1}{3}\ln 2 - \int\limits_0^1 {\left( {({x^2} - x + 1) - \frac{1}{{x + 1}}} \right)dx} } \right]\\
 = \frac{\pi }{{18}}(12\ln 2 - 5).
\end{array}
\end{array}\)