Giả sử tích phân \(I = \int\limits_0^4 {x\ln {{\left( {2x + 1} \right)}^{2017}}dx = a + \dfrac{b}{c}\ln 3.} \) Với phân số \(\dfrac{b}{c}\) tối giản. Lúc đó :
Trả lời bởi giáo viên
Đặt $\left\{ \begin{array}{l}u = \ln {(2x + 1)^{2017}}\\dv = xdx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = \dfrac{{2017.2.{{(2x + 1)}^{2016}}}}{{{{(2x + 1)}^{2017}}}}dx = \dfrac{{4034}}{{2x + 1}}dx\\v = \dfrac{{{x^2}}}{2}\end{array} \right.$
$\left. {I = \ln {{(2x + 1)}^{2017}}.\dfrac{{{x^2}}}{2}} \right|_0^4 - \int_0^4 {\dfrac{{{x^2}}}{2}.\dfrac{{4034}}{{2x + 1}}dx} $
$= \ln {(2.4 + 1)^{2017}}.\dfrac{{{4^2}}}{2} - 0 - 2017\int_0^4 {\dfrac{{{x^2}}}{{2x + 1}}dx}$
$ = 8\ln {9^{2017}} - 2017\int_0^4 {(\dfrac{1}{2}x - \dfrac{1}{4} + \dfrac{{\dfrac{1}{4}}}{{2{\rm{x}} + 1}})dx} $
$= 8\ln {9^{2017}} - \dfrac{{2017}}{2}.\left. {\dfrac{{{x^2}}}{2}} \right|_0^4 + \dfrac{{2017}}{4}\left. x \right|_0^4 - \dfrac{{2017}}{4}\int_0^4 {\dfrac{1}{2}.\dfrac{1}{{2x + 1}}d(2x + 1)}$
$\begin{array}{l} = 8\ln {9^{2017}} - \dfrac{{2017}}{4}{.4^2} + \dfrac{{2017}}{4}.4 - \dfrac{{2017}}{8}\left. {\ln \left| {2x + 1} \right|} \right|_0^4 \\ = 8\ln {9^{2017}} - 6051 - \dfrac{{2017}}{8}.(\ln 9 - \ln 1)\\ = 8\ln {9^{2017}} - 6051 - \dfrac{{2017}}{8}.\ln 9 = \dfrac{{127071}}{4}.\ln 3 - 6051\end{array}$
$ \Rightarrow b + c = 127075$
Hướng dẫn giải:
- Bước 1: Đặt \(\left\{ \begin{array}{l}u = \ln \left( {ax + b} \right)\\dv = f\left( x \right)dx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = \dfrac{a}{{\left( {ax + b} \right)}}dx\\v = \int {f\left( x \right)dx} \end{array} \right.\)
- Bước 2: Tính tích phân theo công thức \(\int\limits_m^n {f\left( x \right)\ln \left( {ax + b} \right)dx} = \left. {uv} \right|_m^n - \int\limits_m^n {vdu} \)