Cho giá trị của tích phân \...

D

- Hướng dẫn giải

\({I_1} = \int\limits_{ - \frac{\pi }{3}}^{\frac{{2\pi }}{3}} {\left( {\in 3x + \cos 3x} \right)dx} = \left. {\left( { - \frac{1}{3}\cos 3x + \frac{1}{3}\sin 3x} \right)} \right|_{ - \frac{\pi }{3}}^{\frac{{2\pi }}{3}} = - \frac{2}{3} \Rightarrow a = - \frac{2}{3}\)

\(\begin{array}{l}{I_2} = \int\limits_e^{2e} {\left( {\frac{1}{x} + \frac{1}{{{x^2}}} - \frac{1}{{x + 1}}} \right)dx} \\ = \left. {\left( {\ln \left| x \right| - \frac{1}{x} - \ln \left| {x + 1} \right|} \right)} \right|_e^{2e}\\ = \ln 2 - \frac{1}{{2e}} + \frac{1}{e} - \ln \left( {2e + 1} \right) + \ln \left( {e + 1} \right)\\ \Rightarrow b = - \frac{1}{{2e}} + \frac{1}{e} + \ln 2 - \ln \left( {2e + 1} \right) + \ln \left( {e + 1} \right)\end{array}\)

\( \Rightarrow a.b \approx - 0,2198\)