Cho hai hàm f(x) và g(x) có đạo hàm trên [1; 2021] thỏa mãn f(2021)

Chọn D.

Ta có $\frac{x}{{\left(x+1\right)}^2}g\left(x\right)+2020x=\left(x+1\right)f\text{'}\left(x\right)\Rightarrow \frac{1}{{\left(x+1\right)}^2}g\left(x\right)-\frac{x+1}{x}f\text{'}\left(x\right)=-2020\left(1\right).$

Mặt khác $\frac{x^3}{x+1}g\text{'}\left(x\right)+f\left(x\right)=2021x^2\Rightarrow \frac{x}{x+1}g\text{'}\left(x\right)+\frac{1}{x^2}.f\left(x\right)=2021\left(2\right).$

Cộng vế theo vế (1) và (2), ta được $\left[\frac{1}{{\left(x+1\right)}^2}g\left(x\right)+\frac{x}{x+1}g\text{'}\left(x\right)\right]-\left[\frac{x+1}{x}f\text{'}\left(x\right)-\frac{1}{x^2}f\left(x\right)\right]=1$

$\iff \left[\frac{x}{x+1}g\left(x\right)-\frac{x+1}{x}f\left(x\right)\right]\text{'}=1\left(*\right).$

Lấy nguyên hàm hai vế (*), ta được $\frac{x}{x+1}g\left(x\right)-\frac{x+1}{x}f\left(x\right)=x+C.$

Vì f(2021) = g(2021) = 0 nên $0=2021+C\iff C=-2021.$

Suy ra $\frac{x}{x+1}g\left(x\right)-\frac{x+1}{x}f\left(x\right)=x-2021$.

Vậy $\underset{1}{\overset{2021}{\int }}\left[\frac{x}{x+1}g\left(x\right)-\frac{x+1}{x}f\left(x\right)\right]dx=\underset{1}{\overset{2021}{\int }}\left(x-2021\right)dx=\left(\frac{1}{2}{x}^2-2021x\right)\left|\begin{array}{l}2021\\ 1\end{array}\right.$

$=-\frac{1}{2}{.2021}^2+2021-\frac{1}{2}.$