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Ta có:
Đặt $u = \ln \left( {1 + x} \right)\,\,\, \Rightarrow \,\,\,du = \frac{{dx}}{{1 + x}}$
$dv = dx\,\,\,\,\, \Rightarrow \,\,\,v = x$
$\begin{array}{l}
I = [\left. {x\ln \left( {1 + x} \right)} \right]_1^2 - \int\limits_1^2 {\frac{x}{{1 + x}}dx} = 2\ln 3 -
\ln 2 - \int\limits_1^2 {\left( {1 - \frac{1}{{1 + x}}} \right)} dx\\
\,\,\, = 2\ln 3 - \ln 2 - \left[ {x - \ln \left| {1 + x} \right|} \right] = 3\ln 3 - 2\ln 2 - 1
\end{array}$
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