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. $\tan x + C$