Hướng dẫn giải

Xét \[U = \int {\frac{{{{\left( {x - 2} \right)}^{2020}}}}{{{{\left( {x + 1} \right)}^{2022}}}}dx} = \int {{{\left( {\frac{{x - 2}}{{x + 1}}} \right)}^{2020}}\frac{1}{{{{\left( {x + 1} \right)}^2}}}dx} \]

Đặt \[u = \frac{{x - 2}}{{x + 1}} \Rightarrow du = \frac{3}{{{{\left( {x + 1} \right)}^2}}}dx \Rightarrow \frac{1}{3}du = \frac{1}{{{{\left( {x + 1} \right)}^2}}}dx\].

Suy ra. \[U = \frac{1}{3}\int {{u^{2020}}du} = \frac{1}{{6063}}{u^{2021}} + C\]. Vậy \[U = \frac{1}{{6063}}{\left( {\frac{{x - 2}}{{x + 1}}} \right)^{2021}} + C\]

Chọn C.