Biết \[\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{4}} {\left( {2{{\cot }^2}x + 5} \right)dx} = \frac{\pi }{a} + b\sqrt 3 + c\] \(\left( {a,{\mkern 1mu} {\mkern 1mu} b,c \in \mathbb{Z}} \right)\). Khi đó giá trị của \(P = a + b + c\) là     

Đáp án đúng là: B

Ta có \(\int\limits_4^8 {f\left( x \right)} dx = \int\limits_4^1 {f\left( x \right)dx} + \int\limits_1^8 {f\left( x \right)dx} \) \( = - \int\limits_1^4 {f\left( x \right)dx} + \int\limits_1^8 {f\left( x \right)dx} = - 3 - 2 = - 5\).

Mặt khác: \(\int\limits_1^4 {\left[ {4f\left( x \right) - 2g\left( x \right)} \right]dx = 4\int\limits_1^4 {f\left( x \right)dx} } - 2\int\limits_1^4 {g\left( x \right)dx} = 4.3 - 2.7 = - 2\).

Ta có \(\int\limits_1^4 {\left[ {f\left( x \right) + g\left( x \right)} \right]} dx = \int\limits_1^4 {f\left( x \right)} dx + \int\limits_1^4 {g\left( x \right)} dx = 3 + 7 = 10\).

Đáp án đúng là: A

Ta có $\underset{1}{\overset{3}{\int }}\frac{x+2}{x}dx=\underset{1}{\overset{3}{\int }}\left(1+\frac{2}{x}\right)dx=\underset{1}{\overset{3}{\int }}dx+\underset{1}{\overset{3}{\int }}\frac{2}{x}dx=2+2{\left.\ln \left|x\right|\right|}_1^3=2+2\ln 3.$

Do đó \(a = 2,\,b = 2,\,c = 3 \Rightarrow S = 7.\)