Đặt \(\left\{{}\begin{matrix}u=3+lnx\\dv=\dfrac{1}{\left(x+1\right)^2}dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x}\\v=-\dfrac{1}{x+1}\end{matrix}\right.\)
\(\Rightarrow I=-\dfrac{3+lnx}{x+1}|^1_3+\int\dfrac{dx}{x\left(x+1\right)}=\left(-\dfrac{3+lnx}{x+1}+ln\left(\dfrac{x}{x+1}\right)\right)|^1_3=...\)



