Tich phan (lnx - 1)dx/(x^2 - ln^2x)
can tu (1) den (e)
\[\begin{array}{l}
\frac{{\ln x - 1}}{{{x^2} - {{\ln }^2}x}} = \frac{{\ln x - 1}}{{\left( {x - \ln x} \right)\left( {x + \ln x} \right)}} = \frac{{\ln x - 1}}{{2x\left( {x - \ln x} \right)}} + \frac{{\ln x - 1}}{{2x\left( {x + \ln x} \right)}}\\
\to I = \int {\left( {\frac{{\ln x - 1}}{{2x\left( {x - \ln x} \right)}} + \frac{{\ln x - 1}}{{2x\left( {x + \ln x} \right)}}} \right)} dx = \int {\left( {\frac{{\ln x - 1}}{{2{x^2}\left( {1 - \frac{{\ln x}}{x}} \right)}} + \frac{{\ln x - 1}}{{2{x^2}\left( {1 + \frac{{\ln x}}{x}} \right)}}} \right)} dx\\
t = \frac{{\ln x}}{x} \to dt = \frac{{1 - \ln x}}{{{x^2}}}dx\\
\left\{ \begin{array}{l}
x = 1 \to t = 0\\
x = e \to t = \frac{1}{e}
\end{array} \right.\\
\to I = \int {\left( {\frac{{ - 1}}{{2\left( {1 - t} \right)}} + \frac{{ - 1}}{{2\left( {1 + t} \right)}}} \right)dt = ...}
\end{array}\]
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