Lời giải:

 \( I = \int_1^e \frac{\ln x - 1}{x^2 \left[ \left( \frac{\ln x}{x} \right)^2 - 1 \right]} \, dx \)

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

 \(\begin{cases}
x = 1 \Rightarrow t = 0   \\
x = e  \Rightarrow t = \frac{1}{e}
\end{cases}\)

\( I = - \int_0^{\frac{1}{2}} \frac{1}{t^2 - 1} \, dt \)

Chú ý !  \( I = \int_0^{\frac{\pi}{2}} \frac{1}{3 \sin^2 x + \cos^2 x} \, dx \)

Không thế đặt: \( t = \tan x \), hoặc \( t = \cot x \)

\( I = \int_0^{\frac{\pi}{4}} \frac{1}{3\sin^2 x  + \cos^2 x } \, dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{1}{3\sin^2 x  + \cos^2 x } \, dx \)

Đặt  \( t = \tan x \):  
\( \Rightarrow I_1 = \int_0^{\frac{\pi}{4}} \frac{1}{\cos^2 x \left[ 3 \tan^2 x + 1 \right]} \, dx \)

Đặt \( t = \cot x \):  
\(\Rightarrow  I_2 = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{1}{\sin^2 x \left[ 3 + \cot^2 x \right]} \, dx \)

\( \Leftrightarrow I_1 + I2 =  \frac{\pi \sqrt{3}}{6} \)