Giả sử tích phân I = tích phân từ 0 đến 4 x ln ( 2 x + 1 )^2017 d x = a + b/c ln 3. . Với phân số b/c tối giản. Lúc đó :

Đặt

\(\left\{ {\begin{array}{*{20}{c}}{u = ln{{(2x + 1)}^{2017}}}\\{dv = xdx}\end{array}} \right.\)

\( \Rightarrow \left\{ {\begin{array}{*{20}{c}}{du = \frac{{2017.2.{{(2x + 1)}^{2016}}}}{{{{(2x + 1)}^{2017}}}}dx = \frac{{4034}}{{2x + 1}}dx}\\{v = \frac{{{x^2}}}{2}}\end{array}} \right.\)

\(I = ln{(2x + 1)^{2017}}.\frac{{{x^2}}}{2}\left| {_0^4} \right. - \int\limits_0^4 {\frac{{{x^2}}}{2}.\frac{{4034}}{{2x + 1}}dx} \)

\[ = \ln {(2.4 + 1)^{2017}}.\frac{{{4^2}}}{2} - 0 - 2017\mathop \smallint \nolimits_0^4 \frac{{{x^2}}}{{2x + 1}}dx\]

\[ = 8\ln {9^{2017}} - 2017\mathop \smallint \nolimits_0^4 (\frac{1}{2}x - \frac{1}{4} + \frac{{\frac{1}{4}}}{{2{\rm{x}} + 1}})dx\]

\[ = 8ln{9^{2017}} - \frac{{2017}}{2}.\frac{{{x^2}}}{2}\left| {_0^4} \right. + \frac{{2017}}{4}x\left| {_0^4} \right. - \frac{{2017}}{4}\int\limits_0^4 {\frac{1}{2}.\frac{1}{{2x + 1}}d(2x + 1)} \]

\[ = 8ln{9^{2017}} - \frac{{2017}}{4}{.4^2} + \frac{{2017}}{4}4 - \frac{{2017}}{8}ln|2x + 1|\left| {_0^4} \right.\]

\[ = 8ln{9^{2017}} - 6051 - \frac{{2017}}{8}.(ln9 - ln1)\]

\[ = 8ln{9^{2017}} - 6051 - \frac{{2017}}{8}.ln9 = \frac{{127071}}{4}.ln3 - 6051\]

\[ \Rightarrow b + c = 127075\]

Đáp án cần chọn là: A