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Câu hỏi: Họ nguyên hàm của hàm số $y={{\left( 2x+1 \right)}^{2019}}$ là
A. $\dfrac{{{\left( 2x+1 \right)}^{2020}}}{4040}+C$.
B. $\dfrac{{{\left( 2x+1 \right)}^{2020}}}{2020}+C$.
C. $\dfrac{{{\left( 2x+1 \right)}^{2018}}}{4036}+C$.
D. $\dfrac{{{\left( 2x+1 \right)}^{2020}}}{2018}+C$.
A. $\dfrac{{{\left( 2x+1 \right)}^{2020}}}{4040}+C$.
B. $\dfrac{{{\left( 2x+1 \right)}^{2020}}}{2020}+C$.
C. $\dfrac{{{\left( 2x+1 \right)}^{2018}}}{4036}+C$.
D. $\dfrac{{{\left( 2x+1 \right)}^{2020}}}{2018}+C$.
Ta có: $I=\int{{{\left( 2x+1 \right)}^{2019}}dx}=\dfrac{1}{2}\int{{{\left( 2x+1 \right)}^{2019}}d\left( 2x+1 \right)=\dfrac{1}{2}.\dfrac{{{\left( 2x+1 \right)}^{2020}}}{2020}+C=\dfrac{{{\left( 2x+1 \right)}^{2020}}}{4040}+C}$.
Đáp án A.