Nếu đặt \(t=\sqrt{3\tan x+1}\) thì tích \(I=\int\limits_{0}^{\frac{\pi }{4}}{\frac{6\tan x}{{{\cos }^{2}}x\sqrt{3\tan x+1}}dx}\) trở thành:
A. \(I=\int\limits_{1}^{2}{\frac{4\left( {{t}^{2}}-1 \right)}{3}dt}\)
B. \(I=\int\limits_{1}^{2}{\left( {{t}^{2}}-1 \right)dt}\)
C. \(\int\limits_{1}^{2}{\frac{\left( {{t}^{2}}-1 \right)}{3}dt}\)
D. \(I=\int\limits_{1}^{2}{\frac{4\left( {{t}^{2}}-1 \right)}{5}}dt\)