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