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Câu hỏi: Nếu $\int{f(x)dx}=\dfrac{1}{x}+\ln x+C$ thì $f(x)$ là
A. $f(x)=-\dfrac{1}{{{x}^{2}}}+\ln x$.
B. $f(x)=\dfrac{x+1}{{{x}^{2}}}$.
C. $f(x)=\sqrt{x}+\ln x$.
D. $f(x)=\dfrac{x-1}{{{x}^{2}}}$.
A. $f(x)=-\dfrac{1}{{{x}^{2}}}+\ln x$.
B. $f(x)=\dfrac{x+1}{{{x}^{2}}}$.
C. $f(x)=\sqrt{x}+\ln x$.
D. $f(x)=\dfrac{x-1}{{{x}^{2}}}$.
Ta có: $\left( \dfrac{1}{x}+\ln x \right)'=-\dfrac{1}{{{x}^{2}}}+\dfrac{1}{x}=\dfrac{x-1}{{{x}^{2}}}$
Vậy $f(x)=\dfrac{x-1}{{{x}^{2}}}$
Vậy $f(x)=\dfrac{x-1}{{{x}^{2}}}$
Đáp án D.