Cho hàm số \[f\left( x \right)\] có \[f\left( 1 \right) = 1\] và \[f'\left( x \right) = - \frac{{\ln x}}{{{x^2}}},\forall x > 0\]. Khi đó \[\int\limits_1^e {f\left( x \right)dx} \] bằng:

Đáp án A

Ta có \[f'\left( x \right) = - \frac{{\ln x}}{{{x^2}}},\forall x > 0 \Rightarrow f\left( x \right) = \int {\left( { - \frac{{\ln x}}{{{x^2}}}} \right)dx} = \int {\ln x.d\left( {\frac{1}{x}} \right)} \]

\[ \Leftrightarrow f\left( x \right) = \frac{{\ln x}}{x} - \int {\frac{1}{{{x^2}}}dx} \Leftrightarrow f\left( x \right) = \frac{{\ln x}}{x} + \frac{1}{x} + C\].

Vì \[f\left( 1 \right) = 1 \Rightarrow C = 0\] nên \[f\left( x \right) = \frac{{\ln x}}{x} + \frac{1}{x}\].

Suy ra \[\int\limits_1^e {f\left( x \right)dx} = \int\limits_1^e {\left( {\frac{{\ln x}}{x} + \frac{1}{x}} \right)dx} = \left( {\frac{{{{\ln }^2}x}}{2} + \ln x} \right)\left| \begin{array}{l}^e\\_1\end{array} \right. = \frac{3}{2}\].

Vậy \[\int\limits_1^e {f\left( x \right)dx} = \frac{3}{2}\].