Hướng dẫn giải

Ta có \(\int\limits_{\frac{\pi }{4}}^{\frac{\pi }{3}} {\frac{{{{\cos }^2}x + \sin x.\cos x + 1}}{{{{\cos }^4}x + \sin x.{{\cos }^3}x}}dx} = \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{3}} {\frac{{2{{\cos }^2}x + \sin x.\cos x + {{\sin }^2}x}}{{{{\cos }^2}x\left( {{{\cos }^2}x + \sin x.\cos x} \right)}}dx} \)

\( = \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{3}} {\frac{{2 + \tan x + {{\tan }^2}x}}{{{{\cos }^2}x\left( {1 + \tan x} \right)}}dx = \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{3}} {\frac{{2 + \tan x + {{\tan }^2}x}}{{\left( {1 + \tan x} \right)}}} d\left( {\tan x} \right)} \)

\( = \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{3}} {\left( {\tan x + \frac{2}{{\left( {1 + \tan x} \right)}}} \right)d\left( {\tan x} \right) = \frac{{{{\tan }^2}x}}{2}} \left| {_{\scriptstyle\atop\scriptstyle\frac{\pi }{4}}^{\scriptstyle\frac{\pi }{3}\atop\scriptstyle}} \right. + 2\ln \left| {\tan x + 1} \right|_{\frac{\pi }{4}}^{\frac{\pi }{3}}\)

\( = 1 - 2\ln 2 + 2\ln \left( {\sqrt 3 + 1} \right).\) Suy ra \(a = 1,b = - 2,c = 2\) nên \(abc = - 4.\)

Chọn C.