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-Đặt $u = {\ln ^2}x\,\,\,\,\, \Rightarrow \,\,\,\,du = \frac{{2\ln x\,dx}}{x}$
$dv = \frac{{dx}}{{{x^2}}}\,\,\,\,\, \Rightarrow \,\,\,\,\,v = \frac{{ - 1}}{x}$
${J_{\left( t \right)}} = - \left. {\frac{1}{x}{{\ln }^2}x} \right]_1^t + \,\,2\int\limits_1^t
{\frac{{\ln x}}{{{x^2}}}dx = - \frac{1}{t}{{\ln }^2}t + 2} \int\limits_1^t {\frac{{\ln
x}}{{{x^2}}}dx} $
- Đặt $u = \ln x\,\,\, \Rightarrow \,\,\,du = \frac{{dx}}{x}$
$dv = \frac{{dx}}{{{x^2}}}\,\,\,\,\, \Rightarrow \,\,\,\,\,v = \frac{{ - 1}}{x}$
Suy ra:$\int\limits_1^t {\frac{{\ln x}}{{{x^2}}}dx} = - \left. {\frac{1}{x}\ln x}
\right]_1^t + \int\limits_1^t {\frac{1}{{{x^2}}}dx} $
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - \left. {\frac{1}{t}\ln t - \frac{1}{x}} \right]_1^t = -
\frac{1}{t}\ln t - \frac{1}{t} + 1$
Vậy $J\left( t \right) = 2 - \frac{1}{t}\left| {{{\left( {\ln t + 1} \right)}^2} + 1} \right| \left( 1
\right)$
Từ ($1$),suy ra $J\left( t \right) < 2$ với $\forall t > 1$
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Đăng bài 09-05-12 09:48 AM
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