Trong phương pháp nguyên hàm từng phần, nếu \(\left\{ {\begin{array}{*{20}{c}}{u = g\left( x \right)}\\{dv = h\left( x \right)dx}\end{array}} \right.\) thì:

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

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

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.−1  

B.3

C.11

D.2