ĐGNL ĐHQG Hà Nội - Tư duy định lượng - Sử dụng phương pháp đổi biến số để tìm nguyên hàm

A.\[dt = u'\left( x \right)dx\]

B. \[dx = u'\left( t \right)dt\]

C. \[dt = \frac{1}{{u\left( x \right)}}dx\]

D. \[dx = \frac{1}{{u\left( t \right)}}dt\]

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

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.\[f\left( x \right)dx = - tdt\]

B. \[f\left( x \right)dx = 2tdt\]

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

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

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.\[F\left( x \right) = - 2\sqrt {1 - \ln x} + \frac{2}{3}\left( {1 - \ln x} \right)\sqrt {1 - \ln x} + 3\]

B. \[F\left( x \right) = - \sqrt {1 - \ln x} + \frac{1}{3}\left( {1 - \ln x} \right)\sqrt {1 - \ln x} + 3\]

C. \[F\left( x \right) = - 2\sqrt {1 - \ln x} - \frac{2}{3}\left( {1 - \ln x} \right)\sqrt {1 - \ln x} + 3\]

D. \[F\left( x \right) = 2\sqrt {1 - \ln x} - \frac{2}{3}\left( {1 - \ln x} \right)\sqrt {1 - \ln x} + 3\]

A.\[I = t - \frac{{{t^2}}}{2} + C\]

B. \[I = \frac{{{t^2}}}{2} - t + C\]

C. \[I = \frac{{{t^2}}}{2} - \frac{{{t^2}}}{3} + C\]

D. \[I = - \frac{{{t^2}}}{2} + \frac{{{t^2}}}{3} + C\]

A.m=2

B.m=−2                  

C.m=1         

D.m=−1

A.6

B.4

C.8

D.5

A.\[I = \frac{4}{3}\smallint \left( {2{u^2} + 1} \right)du\]

B. \[I = \frac{4}{3}\smallint \left( { - {u^2} + 1} \right)du\]

C. \[I = \frac{4}{3}\smallint \left( {{u^2} - 1} \right)du\]

D. \[I = \frac{4}{3}\smallint \left( {2{u^2} - 1} \right)du\]

A.−2

B.2

C.−1

D.1

A.\[dx = \tan tdt\]

B. \[dx = - \left( {1 + {{\cot }^2}t} \right)dt\]

C. \[dx = \left( {1 + {{\tan }^2}t} \right)dt\]

D. \[dx = - \left( {1 + {{\cot }^2}x} \right)dt\]

A.\[f\left( x \right)dx = \left( {1 + {{\tan }^2}t} \right)dt\]

B. \[f\left( x \right)dx = dt\]

C. \[f\left( x \right)dx = \left( {1 + {t^2}} \right)dt\]

D. \[f\left( x \right)dx = \left( {1 + {{\cot }^2}t} \right)dt\]

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

B. \[x = 1\]

C. \[x = - 1\]

D. \[x = 0\]

A.\[\smallint f\left( x \right)d{\rm{x}} = \frac{1}{3}\sqrt {3{{\rm{x}}^2} + 2} + C\]

B. \[\smallint f\left( x \right)d{\rm{x}} = - \frac{1}{3}\sqrt {3{{\rm{x}}^2} + 2} + C\]

C. \[\smallint f\left( x \right)d{\rm{x}} = \frac{1}{6}\sqrt {3{{\rm{x}}^2} + 2} + C\]

D. \[\smallint f\left( x \right)d{\rm{x}} = \frac{2}{3}\sqrt {3{{\rm{x}}^2} + 2} + C\]

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.\[\frac{{{\pi ^2}}}{2} + \ln \frac{\pi }{2} + 1\]

B. \[\frac{{{\pi ^2}}}{4} - \ln \frac{\pi }{2} + 1.\]

C. \[\frac{{{\pi ^2}}}{8}.\]

D. \[\frac{{{\pi ^2}}}{8} + \ln \frac{\pi }{2} + 1.\]Trả lời:

A.3

B.4

C.1

D.2

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

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

C. \[\sin x + C\]

D. \[\tan x + C\]