Đáp án:

• Đặt \( \begin{cases} 
u = \ln x \\ 
dv = \frac{x^2 - 1}{x^2} \, dx = \left(1 - \frac{1}{x^2}\right) \, dx 
\end{cases} \quad \Rightarrow \quad 
\begin{cases} 
du = \frac{1}{x} \, dx \\ 
v = x + \frac{1}{x} 
\end{cases} \)

• \( I = \left(x + \frac{1}{x}\right) \ln x - \int \left(x + \frac{1}{x}\right) \frac{1}{x} \, dx \)

   \(= \left(x + \frac{1}{x}\right) \ln x - \int \left(1 + \frac{1}{x^2}\right) \, dx \)

   \(= \left(x + \frac{1}{x}\right) \ln x - \left(x - \frac{1}{x}\right) + C\)