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Câu hỏi: Tìm họ nguyên hàm của hàm số $y=x^2-3^x+\dfrac{1}{x}$.
A. $\dfrac{x^3}{3}-\dfrac{3^x}{\ln 3}-\dfrac{1}{x^2}+C, C \in \mathbb{R}$.
B. $\dfrac{x^3}{3}-\dfrac{3^x}{\ln 3}-\ln |x|+C, C \in \mathbb{R}$.
C. $\dfrac{x^3}{3}-\dfrac{3^x}{\ln 3}+\ln |x|+C, C \in \mathbb{R}$
D. $\dfrac{x^3}{3}-3^x+\dfrac{1}{x^2}+C, C \in \mathbb{R}$.
A. $\dfrac{x^3}{3}-\dfrac{3^x}{\ln 3}-\dfrac{1}{x^2}+C, C \in \mathbb{R}$.
B. $\dfrac{x^3}{3}-\dfrac{3^x}{\ln 3}-\ln |x|+C, C \in \mathbb{R}$.
C. $\dfrac{x^3}{3}-\dfrac{3^x}{\ln 3}+\ln |x|+C, C \in \mathbb{R}$
D. $\dfrac{x^3}{3}-3^x+\dfrac{1}{x^2}+C, C \in \mathbb{R}$.
Ta có: $\int\left(x^2-3^x+\dfrac{1}{x}\right) \mathrm{d} x=\dfrac{x^3}{3}-\dfrac{3^x}{\ln 3}+\ln |x|+C, C \in \mathbb{R}$.
Đáp án C.