A. \(I = \ln \left( {\frac{{{\pi ^2}}}{4} - 1} \right) - \ln \left( {\frac{{{\pi ^2}}}{{16}} + \frac{{\sqrt 2 }}{2}} \right)\)
B. \(I = \ln \left( {\frac{{{\pi ^2}}}{4} + 1} \right) - \ln \left( {\frac{{{\pi ^2}}}{{16}} + \frac{{\sqrt 2 }}{2}} \right)\)
C. \(I = \ln \left( {\frac{{{\pi ^2}}}{4} - 1} \right) + \ln \left( {\frac{{{\pi ^2}}}{{16}} + \frac{{\sqrt 2 }}{2}} \right)\)
D. \(I = \ln \left( {\frac{{{\pi ^2}}}{4} + 1} \right) + \ln \left( {\frac{{{\pi ^2}}}{{16}} + \frac{{\sqrt 2 }}{2}} \right)\)
B
Đặt \(t = {x^2} + \sin x\)
\(\begin{array}{l} I = \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {\frac{{2x + \cos x}}{{{x^2} + \sin x}}dx} \\ I = \int\limits_{\frac{{{\pi ^2}}}{{16}} + \frac{{\sqrt 2 }}{2}}^{\frac{{{\pi ^2}}}{4} + 1} {\frac{1}{t}dt} \\ = \ln \left( {\frac{{{\pi ^2}}}{4} + 1} \right) - \ln \left( {\frac{{{\pi ^2}}}{{16}} + \frac{{\sqrt 2 }}{2}} \right) \end{array}\)
Câu hỏi trên thuộc đề trắc nghiệm dưới đây !
Copyright © 2021 HOCTAP247