\(\int \limits_1^e \frac{{\ln x}}{{{x^2}}}dx\) bằng?
A. \( - 1 - \frac{1}{e}\)
B. \(1 - \frac{2}{e}\)
C. \( - 1 + \frac{2}{e}\)
D.
Đặt \(dv = \frac{1}{{{x^2}}}dx,u = \ln x\)
Suy ra \(v = - \frac{1}{x},du = \frac{{dx}}{x}\)
Ta có:
\(\begin{array}{l}
\int\limits_1^e {\frac{{\ln x}}{{{x^2}}}} dx = - \left. {\frac{{\ln x}}{x}} \right|_1^e + \int\limits_1^e {\frac{1}{{{x^2}}}dx} = - \frac{1}{e} - \left. {\frac{1}{x}} \right|_1^e\\
= - \frac{1}{e} - \left( {\frac{1}{e} - 1} \right) = 1 - \frac{2}{e}
\end{array}\)
-- Mod Toán 12
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