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Câu hỏi: Tính tích phân $I=\int\limits_{0}^{2}{\dfrac{{{\left( x-2 \right)}^{2018}}}{{{\left( x+1 \right)}^{2020}}}dx}$.
A. $I=\dfrac{{{2}^{2019}}}{3.2020}$.
B. $I=\dfrac{{{2}^{2020}}}{3.2019}$.
C. $I=\dfrac{{{2}^{2019}}}{3.2019}$.
D. $I=\dfrac{{{2}^{2020}}}{3.2021}$.
A. $I=\dfrac{{{2}^{2019}}}{3.2020}$.
B. $I=\dfrac{{{2}^{2020}}}{3.2019}$.
C. $I=\dfrac{{{2}^{2019}}}{3.2019}$.
D. $I=\dfrac{{{2}^{2020}}}{3.2021}$.
Ta có: $I=\int\limits_{0}^{2}{\dfrac{{{\left( x-2 \right)}^{2018}}}{{{\left( x+1 \right)}^{2020}}}dx}=\int\limits_{0}^{2}{{{\left( \dfrac{x-2}{x+1} \right)}^{2018}}.\dfrac{1}{{{\left( x+1 \right)}^{2}}}dx}$.
Đặt: $t=\dfrac{x-2}{x+1}\Rightarrow dt=\dfrac{3}{{{\left( x+1 \right)}^{2}}}dx\Rightarrow \dfrac{1}{3}dt=\dfrac{1}{{{\left( x+1 \right)}^{2}}}dx$.
Đổi cận $x=0\Rightarrow t=-2,x=2\Rightarrow t=0\Rightarrow I=\dfrac{1}{3}\int\limits_{-2}^{0}{{{t}^{2018}}dt}=\left. \dfrac{1}{3}.\dfrac{{{t}^{2019}}}{2019} \right|_{-2}^{0}=\dfrac{{{2}^{2019}}}{3.2019}$.
Đặt: $t=\dfrac{x-2}{x+1}\Rightarrow dt=\dfrac{3}{{{\left( x+1 \right)}^{2}}}dx\Rightarrow \dfrac{1}{3}dt=\dfrac{1}{{{\left( x+1 \right)}^{2}}}dx$.
Đổi cận $x=0\Rightarrow t=-2,x=2\Rightarrow t=0\Rightarrow I=\dfrac{1}{3}\int\limits_{-2}^{0}{{{t}^{2018}}dt}=\left. \dfrac{1}{3}.\dfrac{{{t}^{2019}}}{2019} \right|_{-2}^{0}=\dfrac{{{2}^{2019}}}{3.2019}$.
Đáp án C.