ĐGNL ĐHQG Hà Nội - Tư duy định lượng - Sử dụng phương pháp nguyên hàm từng phần để tìm nguyên hàm

A.\[\smallint udv = uv + \smallint vdu\]

B. \[\smallint udv = uv - \smallint vdu\]

C. \[\smallint udv = \smallint uv - \smallint vdu\]

D. \[\smallint udv = \smallint uvdv - \smallint vdu\]

A.\(\left\{ {\begin{array}{*{20}{c}}{du = g'\left( x \right)dx}\\{v = \smallint h(x)dx}\end{array}} \right.\)

B. \(\left\{ {\begin{array}{*{20}{c}}{du = g\left( x \right)dx}\\{v = \smallint h(x)dx}\end{array}} \right.\)

C. \(\left\{ {\begin{array}{*{20}{c}}{du = \smallint g\left( x \right)dx}\\{v = h(x)dx}\end{array}} \right.\)

D. \(\left\{ {\begin{array}{*{20}{c}}{du = g'\left( x \right)dx}\\{v = h(x)dx}\end{array}} \right.\)

A.\[I = \left( {x + 1} \right)f\left( x \right) - 2F\left( x \right) + C\]

B. \[I = F\left( x \right) - \left( {x + 1} \right)f\left( x \right)\]

C. \[I = \left( {x + 1} \right)f\left( x \right) + C\]

D. \[I = \left( {x + 1} \right)f\left( x \right) - F\left( x \right) + C\]

A.\[\smallint f(x)dx = {x^3}\ln 3x - \frac{{{x^3}}}{3} + C\]

B. \[\smallint f(x)dx = \frac{{{x^3}\ln 3x}}{3} - \frac{{{x^3}}}{9} + C\]

C. \[\smallint f(x)dx = \frac{{{x^3}\ln 3x}}{3} - \frac{{{x^3}}}{3} + C\]

D. \[\smallint f(x)dx = \frac{{{x^3}\ln 3x}}{3} - \frac{{{x^3}}}{{27}} + C\]

A.−1  

B.3

C.11

D.2

A.\[a = 2\]

B. \[a = - 1\]

C. \[a = 0\]

D. \[a = 1\]

A.\[F\left( x \right) = \frac{{{2^x} + \ln 2 - 1}}{{{e^x}\left( {\ln 2 - 1} \right)}}\]

B. \[F\left( x \right) = \frac{1}{{\ln 2 - 1}}{\left( {\frac{2}{e}} \right)^x} + {\left( {\frac{1}{e}} \right)^x} - \frac{1}{{\ln 2 - 1}}\]

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

D. \[F\left( x \right) = {\left( {\frac{2}{e}} \right)^x}\]

A.\[\frac{1}{2}\left( {\frac{1}{4}\sin 2x - \frac{x}{2}\cos 2x} \right) + C\]

B. \[ - \frac{1}{2}\left( {\frac{1}{2}\sin 2x - \frac{x}{4}\cos 2x} \right) + C\]

C. \[\frac{1}{2}\left( {\frac{1}{2}\sin 2x + \frac{x}{2}\cos 2x} \right) + C\]

D. \[ - \frac{1}{2}\left( {\frac{1}{2}\sin 2x + \frac{x}{4}\cos 2x} \right) + C\]

A.\[2\left( {\sqrt x \sin \sqrt x - \cos \sqrt x } \right) + C\]

B. \[2\left( {\sqrt x \sin \sqrt x + \cos \sqrt x } \right) + C\]

C. \[\sqrt x \sin \sqrt x + \cos \sqrt x + C\]

D. \[\sqrt x \sin \sqrt x - \cos \sqrt x + C\]

A.F(x) là hàm chẵn.

B.F(x) là hàm lẻ.

C.F(x) là hàm tuần hoàn với chu kì 2π.

D.F(x) không là hàm chẵn cũng không là hàm lẻ.

A.\[F\left( \pi \right) = - 1\]

B. \[F\left( \pi \right) = \frac{1}{2}\]

C. \[F\left( \pi \right) = 1\]

D. \[F\left( \pi \right) = 0\]

A.\[ - \frac{1}{2}{x^2} + x\tan x + \ln \left| {\cos x} \right| + C\]

B. \[ - \frac{1}{2}{x^2} + x\tan x - \ln \left| {\cos x} \right| + C\]

C. \[\frac{1}{2}{x^2} + x\tan x - \ln \left| {\cos x} \right| + C\]

D. \[\frac{1}{2}{x^2} - x\tan x + \ln \left| {\cos x} \right| + C\]

A.\[I = \frac{1}{2}\left( {1 + \sin 2x} \right)\ln \left( {1 + \sin 2x} \right) - \frac{1}{4}\sin 2x + C\]

B. \[I = \frac{1}{4}\left( {1 + \sin 2x} \right)\ln \left( {1 + \sin 2x} \right) - \frac{1}{2}\sin 2x + C\]

C. \[I = \frac{1}{4}\left( {1 + \sin 2x} \right)\ln \left( {1 + \sin 2x} \right) - \frac{1}{4}\sin 2x + C\]

D. \[I = \frac{1}{4}\left( {1 + \sin 2x} \right)\ln \left( {1 + \sin 2x} \right) + \frac{1}{4}\sin 2x + C\]

A.\[x\ln \left( {x + \sqrt {{x^2} + 1} } \right) - \sqrt {{x^2} + 1} + C\]

B. \[\ln \left( {x + \sqrt {{x^2} + 1} } \right) - \sqrt {{x^2} + 1} + C\]

C. \[x\ln \left( {x + \sqrt {{x^2} + 1} } \right) + \sqrt {{x^2} + 1} + C\]

D. \[\ln \left( {x + \sqrt {{x^2} + 1} } \right) + \sqrt {{x^2} + 1} + C\]

A.\[\frac{{{e^{2x}}}}{{13}}\left( {2\sin 3x + 3\cos 3x} \right) + C\]

B. \[\frac{{{e^{2x}}}}{{13}}\left( {3\sin 3x - 2\cos 3x} \right) + C\]

C. \[\frac{{{e^{2x}}}}{{13}}\left( {2\sin 3x - 3\cos 3x} \right) + C\]

D. \[\frac{{{e^{2x}}}}{{13}}\left( {3\sin 3x + 2\cos 3x} \right) + C\]

A.\[F\left( x \right) = x{e^x} + 1 - \ln \left| {x{e^x} + 1} \right| + C\]

B. \[F\left( x \right) = {e^x} + 1 - \ln \left| {x{e^x} + 1} \right| + C\]

C. \[F\left( x \right) = x{e^x} + 1 - \ln \left| {x{e^{ - x}} + 1} \right| + C\]

D. \[F\left( x \right) = x{e^x} + 1 + \ln \left| {x{e^x} + 1} \right| + C\]

A.\[\frac{x}{{{x^2} + 1}} + C\]

B. \[\frac{{2x}}{{{x^2} + 1}} + C\]

C. \[\frac{{ - x}}{{{x^2} + 1}} + C\]

D. \[\frac{{ - 2x}}{{{x^2} + 1}} + C\]

A.\[\frac{7}{2}\]

B. \[\frac{{5 - e}}{2}\]

C. \[\frac{{7 - e}}{2}\]

D. \[\frac{5}{2}\]

A.\[\left( {x + 1} \right){e^x} + C\]

B. \[\left( {x + 1} \right){e^x} - x + C\]

C. \[\left( {x + 2} \right){e^x} - x + C\]

D. \[\left( {x + 1} \right){e^x} + x + C\]

A.\[2020\left( {x - 2} \right){e^x} + C\]

B. \[x{e^x} + C\]

C. \[2020\left( {x + 2} \right){e^x} + C\]

D. \[\left( {x - 2} \right){e^x} + C\]