Đáp án:

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

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

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

   \( = -\frac{1}{2} (1 + \ln(x+1)) + \ln \left| \frac{x}{x+1} \right| + C \)