Cho nguyên hàm \[I = \smallint \frac{{6tanx}}{{{{\cos }^2}x\sqrt {3\tan x + 1} }}dx\] . Giả sử đặt \[u = \sqrt {3tanx + 1} \;\] thì ta được:

A.\[I = - \,\smallint {\cos ^2}t\,\,{\rm{d}}t.\]

B. \[I = \smallint {\sin ^2}t\,\,{\rm{d}}t.\]

C. \[I = \smallint {\cos ^2}t\,\,{\rm{d}}t.\]

D. \[I = \frac{1}{2}\smallint \left( {1 + \cos 2t} \right){\rm{d}}t.\]

A.\[\ln \left| {\cos x} \right| + C\]

B. \[\ln \left| {\sin x} \right| + C\]

C. \[\sin x + C\]

D. \[\tan x + C\]

A.\[xf\left( {{x^2}} \right)dx = f\left( t \right)dt\]

B. \[xf\left( {{x^2}} \right)dx = \frac{1}{2}f\left( t \right)dt\]

C. \[xf\left( {{x^2}} \right)dx = 2f\left( t \right)dt\]

D. \[xf\left( {{x^2}} \right)dx = {f^2}\left( t \right)dt\]

A.\[I = \frac{1}{5}{\left( {{x^3} + 1} \right)^2}\sqrt {{x^3} + 1} - \frac{1}{3}\left( {{x^3} + 1} \right)\sqrt {{x^3} + 1} + C\]

B. \[I = \frac{2}{5}{\left( {{x^3} + 1} \right)^2}\sqrt {{x^3} + 1} - \frac{2}{3}\left( {{x^3} + 1} \right)\sqrt {{x^3} + 1} + C\]

C. \[I = \frac{2}{5}{\left( {{x^3} + 1} \right)^2}\sqrt {{x^3} + 1} + C\]

D. \[I = \frac{2}{5}{\left( {{x^3} + 1} \right)^2}\sqrt {{x^3} + 1} + \left( {{x^3} + 1} \right)\sqrt {{x^3} + 1} + C\]

A.\[x = 1 - \sqrt 3 \]

B. \[x = 1\]

C. \[x = - 1\]

D. \[x = 0\]

A.\[\smallint f\left( {3x} \right){\rm{d}}x = 2x\ln \left( {9x - 1} \right) + C.\]

B. \[\smallint f\left( {3x} \right){\rm{d}}x = 6x\ln \left( {3x - 1} \right) + C.\]

C. \[\smallint f\left( {3x} \right){\rm{d}}x = 6x\ln \left( {9x - 1} \right) + C.\]

D. \[\smallint f\left( {3x} \right){\rm{d}}x = 3x\ln \left( {9x - 1} \right) + C.\]