Cho tích phân \(I = \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{6\tan x}}{{{{\cos }^2}x\sqrt {3\tan x + 1} }}dx} \). Đặt \(u = \sqrt {3\tan x + 1} \), mệnh đề nào dưới đây đúng?
A.\(I = \dfrac{4}{3}\int\limits_0^{\dfrac{\pi }{4}} {\left( {2{u^2} - 1} \right)du} \)
B.\(I = \dfrac{2}{3}\int\limits_0^{\dfrac{\pi }{4}} {\left( {2{u^2} + 1} \right)du} \)
C.\(I = \dfrac{2}{3}\int\limits_1^2 {\left( {2{u^2} + 1} \right)du} \)
D.\(I = \dfrac{4}{3}\int\limits_1^2 {\left( {{u^2} - 1} \right)du} \)
A.\(I = \dfrac{4}{3}\int\limits_0^{\dfrac{\pi }{4}} {\left( {2{u^2} - 1} \right)du} \)
B.\(I = \dfrac{2}{3}\int\limits_0^{\dfrac{\pi }{4}} {\left( {2{u^2} + 1} \right)du} \)
C.\(I = \dfrac{2}{3}\int\limits_1^2 {\left( {2{u^2} + 1} \right)du} \)
D.\(I = \dfrac{4}{3}\int\limits_1^2 {\left( {{u^2} - 1} \right)du} \)
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Hóa Học lớp 12
