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Câu hỏi: Nếu $\int\limits_{{}}^{{}}{f\left( x \right)dx}=\dfrac{1}{x}+\ln \left| 2x \right|+C$ thì hàm số $f\left( x \right)$ là
A. $f\left( x \right)=\dfrac{-1}{{{x}^{2}}}+\dfrac{1}{x}$
B. $f\left( x \right)=\dfrac{1}{{{x}^{2}}}+\ln \left( 2x \right)$
C. $f\left( x \right)=\sqrt{x}+\dfrac{1}{2x}$
D. $f\left( x \right)=\dfrac{-1}{{{x}^{2}}}+\dfrac{1}{2x}$
A. $f\left( x \right)=\dfrac{-1}{{{x}^{2}}}+\dfrac{1}{x}$
B. $f\left( x \right)=\dfrac{1}{{{x}^{2}}}+\ln \left( 2x \right)$
C. $f\left( x \right)=\sqrt{x}+\dfrac{1}{2x}$
D. $f\left( x \right)=\dfrac{-1}{{{x}^{2}}}+\dfrac{1}{2x}$
Phương pháp:
Sử dụng: $f\left( x \right)={{\left( \int\limits_{{}}^{{}}{f\left( x \right)dx} \right)}^{\prime }}$.
Cách giải:
Ta có: $\int\limits_{{}}^{{}}{f\left( x \right)dx}=\dfrac{1}{x}+\ln \left| 2x \right|+C\Rightarrow f\left( x \right)={{\left( \int\limits_{{}}^{{}}{f\left( x \right)dx} \right)}^{\prime }}=-\dfrac{1}{{{x}^{2}}}+\dfrac{1}{2x}.$
Sử dụng: $f\left( x \right)={{\left( \int\limits_{{}}^{{}}{f\left( x \right)dx} \right)}^{\prime }}$.
Cách giải:
Ta có: $\int\limits_{{}}^{{}}{f\left( x \right)dx}=\dfrac{1}{x}+\ln \left| 2x \right|+C\Rightarrow f\left( x \right)={{\left( \int\limits_{{}}^{{}}{f\left( x \right)dx} \right)}^{\prime }}=-\dfrac{1}{{{x}^{2}}}+\dfrac{1}{2x}.$
Đáp án D.