Câu 4 Ta có: \ I = \int_1^2 {\frac{{{x^2} + x + 2\ln x}}{{{{\left( {x + 2} \right)}^2}}}} dx = \int_1^2 {\frac{{{{\left( {x + 2} \right)}^2} - 3\left( {x + 2} \right) + 2 + 2\ln x}}{{{{\left( {x + 2} \right)}^2}}}} dx.
\ = \int_1^2 {\left[ {1 - \frac{3}{{x + 2}} + \frac{2}{{{{\left( {x + 2} \right)}^2}}} + \frac{{2\ln x}}{{{{\left( {x + 2} \right)}^2}}}} \right]} dx = \left( {x - 3\ln \left| {x + 2} \right| - \frac{2}{{x + 2}}} \right)\left| \begin{array}{l}
2\\
1
\end{array} \right. + 2J.
\ = \frac{{19}}{6} - 6\ln 2 + 3\ln 3 + 2J\left( {J = \int_1^2 {\frac{{\ln x}}{{{{\left( {x + 2} \right)}^2}}}dx} } \right).
- Tính J: Đặt \ \left\{ \begin{array}{l}
u = \ln x\\
dv = \frac{{dx}}{{{{\left( {x + 2} \right)}^2}}}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
du = \frac{{dx}}{x}\\
v = - \frac{1}{{x + 2}}
\end{array} \right.
\ \Rightarrow J = - \frac{{\ln x}}{{x + 2}}\left| \begin{array}{l}
2\\
1
\end{array} \right. + \frac{1}{2}\ln \left| {\frac{x}{{x + 2}}} \right|\left| \begin{array}{l}
2\\
1
\end{array} \right. = \frac{1}{2}\ln 3 - \frac{3}{4}\ln 2 \Rightarrow I = \frac{{19}}{6} - \frac{{15}}{2}\ln 2 + 4\ln 3.
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