Câu hỏi: Họ nguyên hàm của hàm số $y={{\left( 2x+1 \right)}^{2020}}$ là
A. $\dfrac{{{\left( 2x+1 \right)}^{2021}}}{2021}+C.$
B. $\dfrac{{{\left( 2x+1 \right)}^{2021}}}{4040}+C.$
C. $\dfrac{{{\left( 2x+1 \right)}^{2021}}}{4042}+C.$
D. $\dfrac{{{\left( 2x+1 \right)}^{2021}}}{4024}+C$.
A. $\dfrac{{{\left( 2x+1 \right)}^{2021}}}{2021}+C.$
B. $\dfrac{{{\left( 2x+1 \right)}^{2021}}}{4040}+C.$
C. $\dfrac{{{\left( 2x+1 \right)}^{2021}}}{4042}+C.$
D. $\dfrac{{{\left( 2x+1 \right)}^{2021}}}{4024}+C$.
Ta có: $\int\limits_{{}}^{{}}{{{\left( 2x+1 \right)}^{2020}}dx}=\dfrac{1}{2}.\dfrac{{{\left( 2x+1 \right)}^{2021}}}{2021}+C=\dfrac{{{\left( 2x+1 \right)}^{2021}}}{4042}+C.$
Đáp án C.