设t=√（x-1),则x=t^2+1,
dx=2tdt,
原式=∫t*2tdt/(t^2+1-2)
=2∫t^2dt/(t^2-1)
=2∫dt+2∫dt/(t^2-1)
=2t+ln|(t-1)/(t+1)|+C
=2√（x-1)+ln|[√(x-1)-1]-ln|[√(x-1)+1]+C.
一个简单的积分题∫[（x-1）^(1/2)]/(x-2) dx =?没学过sqrt
设t=√（x-1),则x=t^2+1,
dx=2tdt,
原式=∫t*2tdt/(t^2+1-2)
=2∫t^2dt/(t^2-1)
=2∫dt+2∫dt/(t^2-1)
=2t+ln|(t-1)/(t+1)|+C
=2√（x-1)+ln|[√(x-1)-1]-ln|[√(x-1)+1]+C.