Đáp án đúng: A
Giải chi tiết:Đặt \(J = \int\limits_0^{\frac{\pi }{6}} {\dfrac{{{{\cos }^2}xdx}}{{\sin x + \sqrt 3 \cos x}}} \)
Ta có \(I - 3J = \int\limits_0^{\frac{\pi }{6}} {\dfrac{{si{n^2}x - 3{{\cos }^2}x}}{{\sin x + \sqrt 3 \cos x}}dx = } \int\limits_0^{\frac{\pi }{6}} {\left( {\sin x - \sqrt 3 \cos x} \right)dx} \)
\(\begin{array}{l} \Rightarrow I - 3J = \left. {\left( { - \cos x - \sqrt 3 \sin x} \right)} \right|_0^{\frac{\pi }{6}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( { - \dfrac{{\sqrt 3 }}{2} - \dfrac{{\sqrt 3 }}{2}} \right) - \left( { - 1} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 - \sqrt 3 \end{array}\)
Mà \(I + J = \int\limits_0^{\frac{\pi }{6}} {\dfrac{{dx}}{{\sin x + \sqrt 3 \cos x}}} = \dfrac{{\ln 3}}{4}\) (sử dụng MTCT).
Khi đó ta có: \(\left\{ \begin{array}{l}I - 3J = 1 - \sqrt 3 \\I + J = \dfrac{{\ln 3}}{4}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}I - 3J = 1 - \sqrt 3 \\3I + 3J = \dfrac{{3\ln 3}}{4}\end{array} \right. \Rightarrow 4I = 1 - \sqrt 3 + \dfrac{{3\ln 3}}{4}\)
\( \Rightarrow I = \dfrac{{1 - \sqrt 3 }}{4} + \dfrac{{3\ln 3}}{{16}} = \dfrac{1}{{16}}\left( {3\ln 3 + 4 - 4\sqrt 3 } \right)\).
Đồng nhất hệ số ta có \(a = 3,\,\,b = 4,\,\,c = - 4\).
Vậy \(a + b + c = 3\).
Chọn A
