A.Bạn A làm sai bước 1 
B.Bạn A làm sai bước 2
C.Bạn A làm sai bước 3 
D.Bạn A làm hoàn toàn đúng

A.\(\int {\dfrac{{2dt}}{{{{\left( {1 + t} \right)}^2}}}} \)
B.\(\int {\dfrac{{2dt}}{{{{\left( {1 + t} \right)}^3}}}} \)
C.\(\int {\dfrac{{dt}}{{3{{\left( {1 + t} \right)}^2}}}} \)
D.\(\int {\dfrac{{2dt}}{{3{{\left( {1 + t} \right)}^2}}}} \)

A.\(I = 2t + \dfrac{{\cos 4t}}{2} + C\)
B.\(I = 2t + \dfrac{{\sin 8t}}{4} + C\)
C.\(I = 2t - \dfrac{{\cos 4t}}{2} + C\)
D.\(I = 2t - \dfrac{{\sin 4t}}{4} + C\)

A.\(\dfrac{{81t}}{8} - \dfrac{{81}}{{32}}\sin 4t + C\)
B.\(\dfrac{t}{2} - \dfrac{1}{8}\sin 4t + C\)
C.\(t - \dfrac{1}{4}\sin 4t + C\)
D.\(2t - \dfrac{1}{2}\sin 4t + C\)

A.\(\int {f\left( x \right)dx}  = \dfrac{{\arcsin x - x\sqrt {1 - {x^2}} }}{2} + C\)
B.\(\int {f\left( x \right)dx}  = \dfrac{{\arcsin x + x\sqrt {1 - {x^2}} }}{2} + C\)
C.\(\int {f\left( x \right)dx}  = \dfrac{{2\arcsin x - x\sqrt {1 - {x^2}} }}{2} + C\)
D.\(\int {f\left( x \right)dx}  = \dfrac{{2\arcsin x + x\sqrt {1 - {x^2}} }}{2} + 2x\)

A.\(F\left( t \right) =  - \tan t + C\)
B.\(F\left( t \right) =  - \cot t + C\)
C.\(F\left( t \right) = \tan t + C\)
D.\(F\left( t \right) = \cot t + C\)

A.\(\tan x + C\)
B.\(\cot x + C\)
C.\(\arctan x + C\)
D.\({\mathop{\rm arccot}\nolimits} x + C\)

A.\(J = \int {\tan tdt} \)
B.\(J = \int {\tan 2tdt} \)
C.\(J = \dfrac{1}{2}\int {\tan 2tdt} \)
D.\(J = \dfrac{1}{2}\int {{{\tan }^2}} tdt\)

A.\(I = \int {2\left( {1 + \cos 2t} \right)dt} \)
B.\(I = \int {2\left( {1 + \cos 3t} \right)dt} \)
C.\(I = \int {2\left( {4 + \cos 2t} \right)dt} \)
D.\(I = \int {2\left( {1 + 2\cos 2t} \right)dt} \)

A.
B.
C.
D.