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Câu hỏi: Tính tích phân $I=\int\limits_{1}^{2}{\dfrac{{{\left( x+2 \right)}^{2018}}}{{{x}^{2020}}}}dx.$
A. $\dfrac{{{3}^{2018}}-{{2}^{2018}}}{4036}.$
B. $\dfrac{{{3}^{2019}}-{{2}^{2019}}}{4038}.$
C. $\dfrac{{{3}^{2017}}}{4034}-\dfrac{{{2}^{2018}}}{2020}$
D. $\dfrac{{{3}^{2021}}-{{2}^{2021}}}{4040}.$
A. $\dfrac{{{3}^{2018}}-{{2}^{2018}}}{4036}.$
B. $\dfrac{{{3}^{2019}}-{{2}^{2019}}}{4038}.$
C. $\dfrac{{{3}^{2017}}}{4034}-\dfrac{{{2}^{2018}}}{2020}$
D. $\dfrac{{{3}^{2021}}-{{2}^{2021}}}{4040}.$
Ta có $I=\int\limits_{1}^{2}{\dfrac{{{\left( x+2 \right)}^{2018}}}{{{x}^{2020}}}dx}=\int\limits_{1}^{2}{{{\left( 1+\dfrac{2}{x} \right)}^{2018}}.\dfrac{1}{{{x}^{2}}}dx}.$
Đặt $t=1+\dfrac{2}{x}\Rightarrow \left\{ \begin{aligned}
& x=\dfrac{2}{t-1}\Rightarrow dx=-\dfrac{2}{{{\left( 1-t \right)}^{2}}}dt \\
& {{x}^{2}}=\dfrac{4}{{{\left( t-1 \right)}^{2}}} \\
\end{aligned} \right. $. Đổi cận: $ \left\{ \begin{aligned}
& x=1\Rightarrow t=3 \\
& x=2\Rightarrow t=2 \\
\end{aligned} \right.$
Suy ra $I=-\int\limits_{3}^{2}{\dfrac{{{t}^{2018}}.2{{\left( t-1 \right)}^{2}}}{4{{\left( t-1 \right)}^{2}}}dt}=\dfrac{1}{2}\int\limits_{2}^{3}{{{t}^{2018}}dt}=\left. \dfrac{{{t}^{2019}}}{4038} \right|_{2}^{3}=\dfrac{{{3}^{2019}}-{{2}^{2019}}}{4038}.$
Đặt $t=1+\dfrac{2}{x}\Rightarrow \left\{ \begin{aligned}
& x=\dfrac{2}{t-1}\Rightarrow dx=-\dfrac{2}{{{\left( 1-t \right)}^{2}}}dt \\
& {{x}^{2}}=\dfrac{4}{{{\left( t-1 \right)}^{2}}} \\
\end{aligned} \right. $. Đổi cận: $ \left\{ \begin{aligned}
& x=1\Rightarrow t=3 \\
& x=2\Rightarrow t=2 \\
\end{aligned} \right.$
Suy ra $I=-\int\limits_{3}^{2}{\dfrac{{{t}^{2018}}.2{{\left( t-1 \right)}^{2}}}{4{{\left( t-1 \right)}^{2}}}dt}=\dfrac{1}{2}\int\limits_{2}^{3}{{{t}^{2018}}dt}=\left. \dfrac{{{t}^{2019}}}{4038} \right|_{2}^{3}=\dfrac{{{3}^{2019}}-{{2}^{2019}}}{4038}.$
Đáp án B.