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Câu hỏi: Cho hàm số $f\left( x \right)$ có $f\left( 1 \right)=1$ và $f'\left( x \right)=-\dfrac{\ln x}{{{x}^{2}}},\forall x>0$. Khi đó $\int\limits_{1}^{e}{f\left( x \right)dx}$ bằng:
A. $\dfrac{3}{2}.$
B. $\dfrac{2}{e}-1.$
C. $-\dfrac{3}{2}.$
D. $1-\dfrac{2}{e}.$
A. $\dfrac{3}{2}.$
B. $\dfrac{2}{e}-1.$
C. $-\dfrac{3}{2}.$
D. $1-\dfrac{2}{e}.$
Ta có $f'\left( x \right)=-\dfrac{\ln x}{{{x}^{2}}},\forall x>0\Rightarrow f\left( x \right)=\int{\left( -\dfrac{\ln x}{{{x}^{2}}} \right)dx}=\int{\ln x.d\left( \dfrac{1}{x} \right)}$
$\Leftrightarrow f\left( x \right)=\dfrac{\ln x}{x}-\int{\dfrac{1}{{{x}^{2}}}dx}\Leftrightarrow f\left( x \right)=\dfrac{\ln x}{x}+\dfrac{1}{x}+C$.
Vì $f\left( 1 \right)=1\Rightarrow C=0$ nên $f\left( x \right)=\dfrac{\ln x}{x}+\dfrac{1}{x}$.
Suy ra $\int\limits_{1}^{e}{f\left( x \right)dx}=\int\limits_{1}^{e}{\left( \dfrac{\ln x}{x}+\dfrac{1}{x} \right)dx}=\left( \dfrac{{{\ln }^{2}}x}{2}+\ln x \right)\left| \begin{aligned}
& ^{e} \\
& _{1} \\
\end{aligned} \right.=\dfrac{3}{2}$.
Vậy $\int\limits_{1}^{e}{f\left( x \right)dx}=\dfrac{3}{2}$.
$\Leftrightarrow f\left( x \right)=\dfrac{\ln x}{x}-\int{\dfrac{1}{{{x}^{2}}}dx}\Leftrightarrow f\left( x \right)=\dfrac{\ln x}{x}+\dfrac{1}{x}+C$.
Vì $f\left( 1 \right)=1\Rightarrow C=0$ nên $f\left( x \right)=\dfrac{\ln x}{x}+\dfrac{1}{x}$.
Suy ra $\int\limits_{1}^{e}{f\left( x \right)dx}=\int\limits_{1}^{e}{\left( \dfrac{\ln x}{x}+\dfrac{1}{x} \right)dx}=\left( \dfrac{{{\ln }^{2}}x}{2}+\ln x \right)\left| \begin{aligned}
& ^{e} \\
& _{1} \\
\end{aligned} \right.=\dfrac{3}{2}$.
Vậy $\int\limits_{1}^{e}{f\left( x \right)dx}=\dfrac{3}{2}$.
Đáp án A.