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.m=2
B.m=−2
C.m=1
D.m=−1
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.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. \[\tan x + C\]
Biết \[\smallint f\left( u \right)du = F\left( u \right) + C\]. Tìm khẳng định đúng
A.\[\smallint f(5x + 2)dx = 5F(x) + 2 + C\]
B. \[\smallint f(5x + 2)dx = F(5x + 2) + C\]
C. \[\smallint f(5x + 2)dx = \frac{1}{5}F(5x + 2) + C\]
D. \[\smallint f(5x + 2)dx = 5F(5x + 2) + C\]
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