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Câu hỏi: Cho hai hàm $f\left( x \right)$ và $g\left( x \right)$ có đạo hàm trên $\left[ 1;2021 \right],$ thỏa mãn $f\left( 2021 \right)=g\left( 2021 \right)=0$,
$\dfrac{x}{{{\left( x+1 \right)}^{2}}}g\left( x \right)+2020x=\left( x+1 \right){f}'\left( x \right)$ và $\dfrac{{{x}^{3}}}{x+1}{g}'\left( x \right)+f\left( x \right)=2021{{x}^{2}}$ với mọi $x\in \left[ 1;2021 \right]$. Tích phân $\int\limits_{1}^{2021}{\left[ \dfrac{x}{x+1}g\left( x \right)-\dfrac{x+1}{x}f\left( x \right) \right]\text{d}x}$ bằng
A. $\dfrac{1}{2}{{.2021}^{2}}-2021+\dfrac{1}{2}.$
B. $\dfrac{1}{2}{{.2020}^{2}}-2020+\dfrac{1}{2}.$
C. $-\dfrac{1}{2}{{.2020}^{2}}+2020-\dfrac{1}{2}.$
D. $-\dfrac{1}{2}{{.2021}^{2}}+2021-\dfrac{1}{2}.$
$\dfrac{x}{{{\left( x+1 \right)}^{2}}}g\left( x \right)+2020x=\left( x+1 \right){f}'\left( x \right)$ và $\dfrac{{{x}^{3}}}{x+1}{g}'\left( x \right)+f\left( x \right)=2021{{x}^{2}}$ với mọi $x\in \left[ 1;2021 \right]$. Tích phân $\int\limits_{1}^{2021}{\left[ \dfrac{x}{x+1}g\left( x \right)-\dfrac{x+1}{x}f\left( x \right) \right]\text{d}x}$ bằng
A. $\dfrac{1}{2}{{.2021}^{2}}-2021+\dfrac{1}{2}.$
B. $\dfrac{1}{2}{{.2020}^{2}}-2020+\dfrac{1}{2}.$
C. $-\dfrac{1}{2}{{.2020}^{2}}+2020-\dfrac{1}{2}.$
D. $-\dfrac{1}{2}{{.2021}^{2}}+2021-\dfrac{1}{2}.$
Ta có $\dfrac{x}{{{\left( x+1 \right)}^{2}}}g\left( x \right)+2020x=\left( x+1 \right)f'\left( x \right)\Rightarrow \dfrac{1}{{{\left( x+1 \right)}^{2}}}g\left( x \right)-\dfrac{x+1}{x}f'\left( x \right)=-2020\left( 1 \right).$
Mặt khác $\dfrac{{{x}^{3}}}{x+1}g'\left( x \right)+f\left( x \right)=2021{{x}^{2}}\Rightarrow \dfrac{x}{x+1}g'\left( x \right)+\dfrac{1}{{{x}^{2}}}.f\left( x \right)=2021\left( 2 \right).$
Cộng vế theo vế (1) và (2), ta được $\left[ \dfrac{1}{{{\left( x+1 \right)}^{2}}}g\left( x \right)+\dfrac{x}{x+1}g'\left( x \right) \right]-\left[ \dfrac{x+1}{x}f'\left( x \right)-\dfrac{1}{{{x}^{2}}}f\left( x \right) \right]=1$
$\Leftrightarrow \left[ \dfrac{x}{x+1}g\left( x \right)-\dfrac{x+1}{x}f\left( x \right) \right]'=1\left( * \right).$
Lấy nguyên hàm hai vế (*), ta được $\dfrac{x}{x+1}g\left( x \right)-\dfrac{x+1}{x}f\left( x \right)=x+C.$
Vì $f\left( 2021 \right)=g\left( 2021 \right)=0$ nên $0=2021+C\Leftrightarrow C=-2021.$
Suy ra $\dfrac{x}{x+1}g\left( x \right)-\dfrac{x+1}{x}f\left( x \right)=x-2021$.
Vậy $\int\limits_{1}^{2021}{\left[ \dfrac{x}{x+1}g\left( x \right)-\dfrac{x+1}{x}f\left( x \right) \right]dx}=\int\limits_{1}^{2021}{\left( x-2021 \right)dx}=\left( \dfrac{1}{2}{{x}^{2}}-2021x \right)\left| \begin{aligned}
& 2021 \\
& 1 \\
\end{aligned} \right.$
$=-\dfrac{1}{2}{{.2021}^{2}}+2021-\dfrac{1}{2}.$
Mặt khác $\dfrac{{{x}^{3}}}{x+1}g'\left( x \right)+f\left( x \right)=2021{{x}^{2}}\Rightarrow \dfrac{x}{x+1}g'\left( x \right)+\dfrac{1}{{{x}^{2}}}.f\left( x \right)=2021\left( 2 \right).$
Cộng vế theo vế (1) và (2), ta được $\left[ \dfrac{1}{{{\left( x+1 \right)}^{2}}}g\left( x \right)+\dfrac{x}{x+1}g'\left( x \right) \right]-\left[ \dfrac{x+1}{x}f'\left( x \right)-\dfrac{1}{{{x}^{2}}}f\left( x \right) \right]=1$
$\Leftrightarrow \left[ \dfrac{x}{x+1}g\left( x \right)-\dfrac{x+1}{x}f\left( x \right) \right]'=1\left( * \right).$
Lấy nguyên hàm hai vế (*), ta được $\dfrac{x}{x+1}g\left( x \right)-\dfrac{x+1}{x}f\left( x \right)=x+C.$
Vì $f\left( 2021 \right)=g\left( 2021 \right)=0$ nên $0=2021+C\Leftrightarrow C=-2021.$
Suy ra $\dfrac{x}{x+1}g\left( x \right)-\dfrac{x+1}{x}f\left( x \right)=x-2021$.
Vậy $\int\limits_{1}^{2021}{\left[ \dfrac{x}{x+1}g\left( x \right)-\dfrac{x+1}{x}f\left( x \right) \right]dx}=\int\limits_{1}^{2021}{\left( x-2021 \right)dx}=\left( \dfrac{1}{2}{{x}^{2}}-2021x \right)\left| \begin{aligned}
& 2021 \\
& 1 \\
\end{aligned} \right.$
$=-\dfrac{1}{2}{{.2021}^{2}}+2021-\dfrac{1}{2}.$
Đáp án D.