Giả sử tích phân \[I = \mathop \smallint \limits_0^4 x\ln {\left( {2x + 1} \right)^{2017}}dx = a + \frac{b}{c}\ln 3.\].  Với phân số  \(\frac{b}{c}\) tối giản. Lúc đó :

A.\(\frac{1}{2}\)

B. \(\frac{1}{4}\)

C. 1

D. 0

A.S=0

B.S=1

C.\[S = \frac{1}{2}\]

D. \[S = \frac{3}{8}\]

A.\[I = f\left( x \right).g'\left( x \right)\left| {_a^b} \right. - \int\limits_a^b {f'\left( x \right)} .g\left( x \right)dx\]

B. \[I = f\left( x \right).g\left( x \right)\left| {_a^b} \right. - \int\limits_a^b {f\left( x \right)} .g\left( x \right)dx\]

C. \[I = f\left( x \right).g\left( x \right)\left| {_a^b} \right. - \int\limits_a^b {f'\left( x \right)} .g\left( x \right)dx\]

D. \[I = f\left( x \right).g'\left( x \right)\left| {_a^b} \right. - \int\limits_a^b {f\left( x \right)} .g'\left( x \right)dx\]

A.\(\left\{ {\begin{array}{*{20}{c}}{u = x}\\{dv = xcosxdx}\end{array}} \right.\)

B. \(\left\{ {\begin{array}{*{20}{c}}{u = {x^2}}\\{dv = cosxdx}\end{array}} \right.\)

C. \(\left\{ {\begin{array}{*{20}{c}}{u = cosx}\\{dv = {x^2}dx}\end{array}} \right.\)

D. \(\left\{ {\begin{array}{*{20}{c}}{u = {x^2}cosx}\\{dv = dx}\end{array}} \right.\)

A.\[I = \frac{1}{2}\]

B. \[I = \frac{{3{e^2} + 1}}{4}\]

C. \[I = \frac{{{e^2} + 1}}{4}\]

D. \[I = \frac{{{e^2} - 1}}{4}\]

A.\[\frac{{32}}{{15}}\]

B. \[\frac{{86}}{{15}}\]

C. \[\frac{{ - 11}}{{15}}\]

D. \[\frac{{16}}{{15}}\]

A.\[I = 0.\]

B. \[I = - \,1.\]

C. \[I = 1.\]

D. \[I = 2.\]