Gọi F(x) là một nguyên hàm của hàm số $y = x.cosx$ mà F(0)=1. Phát biểu nào sau đây đúng:

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

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

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.$\smallint udv = uv + \smallint vdu$

B. $\smallint udv = uv - \smallint vdu$

C. $\smallint udv = \smallint uv - \smallint vdu$

D. $\smallint udv = \smallint uvdv - \smallint vdu$