Hướng dẫn giải

Ta có \(I = \int\limits_0^{\frac{\pi }{6}} {\frac{{dx}}{{1 + \sin x}} = \int\limits_0^{\frac{\pi }{6}} {\frac{{dx}}{{{{\left( {\cos \frac{x}{2} + \sin \frac{x}{2}} \right)}^2}}} = \int\limits_0^{\frac{\pi }{6}} {\frac{{\frac{1}{{{{\cos }^2}\frac{x}{2}}}}}{{{{\left( {1 + \tan \frac{x}{2}} \right)}^2}}}dx} = \int\limits_0^{\frac{\pi }{6}} {\frac{{\left( {1 + {{\tan }^2}\frac{x}{2}} \right)}}{{{{\left( {1 + \tan \frac{x}{2}} \right)}^2}}}dx.} } } \)

Đặt \(t = 1 + \tan \frac{x}{2} \Rightarrow 2dt = \left( {1 + {{\tan }^2}\frac{x}{2}} \right)dx.\)

Đổi cận \(x = 0 \Rightarrow t = 1;x = \frac{\pi }{6} \Rightarrow t = 3 - \sqrt 3 .\)

\(I = \int\limits_1^{3 - \sqrt 3 } {\frac{{2dt}}{{{t^2}}} = - \frac{2}{t}\left| {_{\scriptstyle\atop\scriptstyle1}^{\scriptstyle3 - \sqrt 3 \atop\scriptstyle}} \right. = \frac{{ - \sqrt 3 + 3}}{3}.} \)

Suy ra \(a = - 1,b = 3,c = 3\) nên \(a + b + c = 5.\)

Chọn A.