Cho I = intlimits_1^e {frac{{ln x - 1}}{{{x^2} - {{ln }^2}x}}dx,} đặt (t = frac{{ln x}}{x}

* Hướng dẫn giải

Đặt: \(t = \frac{{\ln x}}{x} \Rightarrow dt = \frac{{1 - \ln x}}{{{x^2}}}dx\)

Đổi cận: \(\left\{ {\begin{array}{*{20}{l}}
{x = 1 \Rightarrow t = 0}\\
{x = e \Rightarrow t = \frac{1}{e}}
\end{array}} \right.\)

Vậy: 

\(\begin{array}{*{20}{l}}
{I = \int\limits_1^e {\frac{{\ln x - 1}}{{{x^2} - {{\ln }^2}x}}dx}  = \int\limits_1^e {\frac{1}{{1 - \frac{{{{\ln }^2}x}}{{{x^2}}}}}.\frac{{\ln x - 1}}{{{x^2}}}dx} }\\
\begin{array}{l}
 = \int\limits_0^{\frac{1}{e}} {\frac{1}{{1 - {t^2}}}dt}  = \frac{1}{2}\int\limits_0^{\frac{1}{e}} {\left( {\frac{1}{{t - 1}} - \frac{1}{{t + 1}}} \right)dt} \\
 = \frac{1}{2}\ln \left( {\frac{{e - 1}}{{e + 1}}} \right).
\end{array}
\end{array}\)