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