đặt $\sqrt{x-1}=t \Rightarrow x-1 =t^2 \Rightarrow dx =2tdt$
$I=2\int \dfrac{t}{t^2 +1 -2t}dt=2\int \bigg (\dfrac{1}{t-1} +\dfrac{1}{(t-1)^2} \bigg )dt=2\ln|t-1| +2\int \dfrac{d(t-1)}{(t-1)^2}$
$=2\ln|t-1| +2\int (t-1)^{-2}d(t-1) =2\ln|t-1|-\dfrac{2}{t-1}+C$ tự thay cận