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