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Câu hỏi: Tìm họ nguyên hàm của hàm số $y=x^{2020}+2002^x+\dfrac{1}{x}$.
A. $\dfrac{x^{2021}}{2021}+\dfrac{2002^x}{\ln 2002}+\ln |x|+C, C \in \mathbb{R}$.
B. $\dfrac{x^{2020}}{\ln 2020}+2002^x \ln 2002+\ln |x|+C, C \in \mathbb{R}$.
C. $\dfrac{x^{2021}}{2021}+\dfrac{2002^x}{\ln 2002}-\dfrac{1}{x^2}+C, C \in \mathbb{R}$.
D. $\dfrac{x^{2021}}{2021}+\dfrac{2002^{x+1}}{x+1}+\ln x+C, C \in \mathbb{R}$.
A. $\dfrac{x^{2021}}{2021}+\dfrac{2002^x}{\ln 2002}+\ln |x|+C, C \in \mathbb{R}$.
B. $\dfrac{x^{2020}}{\ln 2020}+2002^x \ln 2002+\ln |x|+C, C \in \mathbb{R}$.
C. $\dfrac{x^{2021}}{2021}+\dfrac{2002^x}{\ln 2002}-\dfrac{1}{x^2}+C, C \in \mathbb{R}$.
D. $\dfrac{x^{2021}}{2021}+\dfrac{2002^{x+1}}{x+1}+\ln x+C, C \in \mathbb{R}$.
Áp dụng các công thức nguyên hàm cơ bản:
$\int x^\alpha d x=\dfrac{x^{\alpha+1}}{\alpha+1}+C, C \in \mathbb{R} ; \int a^x d x=\dfrac{a^x}{\ln a}+C, C \in \mathbb{R} ; \int \dfrac{1}{x} d x=\ln |x|+C, C \in \mathbb{R}$.
Ta có: $\int\left(x^{2020}+2002^x+\dfrac{1}{x}\right) \mathrm{d} x=\dfrac{x^{2021}}{2021}+\dfrac{2002^x}{\ln 2002}+\ln |x|+C, C \in \mathbb{R}$.
$\int x^\alpha d x=\dfrac{x^{\alpha+1}}{\alpha+1}+C, C \in \mathbb{R} ; \int a^x d x=\dfrac{a^x}{\ln a}+C, C \in \mathbb{R} ; \int \dfrac{1}{x} d x=\ln |x|+C, C \in \mathbb{R}$.
Ta có: $\int\left(x^{2020}+2002^x+\dfrac{1}{x}\right) \mathrm{d} x=\dfrac{x^{2021}}{2021}+\dfrac{2002^x}{\ln 2002}+\ln |x|+C, C \in \mathbb{R}$.
Đáp án A.