đặt $t=\sqrt{x-1}\Rightarrow 2tdt=dx$đổi cận $ x=2\Rightarrow t=1, x=5\Rightarrow t=2$
I=2$\int\limits_{1}^{2}\frac{ln(t+1)}{t^2+t}tdt=2\int\limits_{1}^{2}\frac{ln(t+1)}{(t+1)}dt=2\int\limits_{1}^{2}ln(t+1) d[ln(t+1)]$
đặt u=ln(t+1) $\Rightarrow du=d[ln(t+1)]$
$t=1\Rightarrow u=ln2,t=2\Rightarrow u=ln3$
$\Rightarrow I=2\int\limits_{ln2}^{ln3}udu=u^2\left| {\begin{matrix} ln3\\ ln2 \end{matrix}} \right.=ln^23-ln^22$