Tính tích phân \(I = \int\limits_4^5 {\left( {x + 1} \right)\ln \left( {x – 3} \right){\rm{d}}x} \)?
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Lời giải:
Báo saiĐặt \(\left\{ \begin{array}{l}u = \ln \left( {x – 3} \right)\\{\rm{d}}v = x + 1\end{array} \right. \Rightarrow \left\{ \begin{array}{l}{\rm{d}}u = \frac{1}{{x – 3}}{\rm{d}}x\\v = \frac{1}{2}{x^2} + x\end{array} \right.\)
\(\begin{array}{l}I = \left( {\frac{1}{2}{x^2} + x} \right)\ln \left( {x – 3} \right)\left| {\begin{array}{*{20}{c}}5\\4\end{array}} \right. – \int\limits_4^5 {\frac{{\frac{1}{2}{x^2} + x}}{{x – 3}}} {\rm{d}}x\\{\mkern 1mu} \end{array}\)
\({\mkern 1mu} {\mkern 1mu} = \frac{{35}}{2}\ln 2 – \frac{1}{2}\int\limits_4^5 {\frac{{{x^2} – 9 + 9}}{{x – 3}}dx – \int\limits_4^5 {\frac{{x – 3 + 3}}{{x – 3}}} } dx\)
\({\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} = \frac{{35}}{2}\ln 2 – \frac{1}{2}\left( {\frac{9}{2} + 3 + 9\ln 2} \right) – \left( {1 + 3\ln 2} \right)\)
\({\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} = 10\ln 2 – \frac{{19}}{4}\)