Nguyên hàm của hàm số f( x ) = x( x + 2 )^2019 là: A. - ( x + 2)^2021/2021- ( x + 2)^2020/1010 + C B. ( x + 2)^2020/2021 - ( x + 2)^2018/1009 + C C. ( x + 2 )^2021/2021 + ( x + 2)^2020/1

7/62

Nguyên hàm của hàm số \[f\left( x \right) = x{\left( {x + 2} \right)^{2019}}\] là:

\[ - \frac{{{{\left( {x + 2} \right)}^{2021}}}}{{2021}} - \frac{{{{\left( {x + 2} \right)}^{2020}}}}{{1010}} + C\]

\[\frac{{{{\left( {x + 2} \right)}^{2020}}}}{{2021}} - \frac{{{{\left( {x + 2} \right)}^{2018}}}}{{1009}} + C\]

\[\frac{{{{\left( {x + 2} \right)}^{2021}}}}{{2021}} + \frac{{{{\left( {x + 2} \right)}^{2020}}}}{{1010}} + C\]

\[\frac{{{{\left( {x + 2} \right)}^{2021}}}}{{2021}} - \frac{{{{\left( {x + 2} \right)}^{2020}}}}{{1010}} + C\]

Giải thích

Hướng dẫn giải

Ta có: \[\begin{array}{l}\int {x{{\left( {x + 2} \right)}^{2019}}dx} = \int {\left[ {\left( {x + 2} \right) - 2} \right]{{\left( {x + 2} \right)}^{2019}}dx} \\ = \int {{{\left( {x + 2} \right)}^{2020}}dx} - 2\int {{{\left( {x + 2} \right)}^{2019}}dx} = \frac{{{{\left( {x + 2} \right)}^{2021}}}}{{2021}} - \frac{{{{\left( {x + 2} \right)}^{2020}}}}{{1010}} + C\end{array}\]

Chọn D.