积分号 1/(1+根号（x-1）)dx 求积分 急，谢谢 详细的过程

不定积分

∫dx/[1+√(x-1)]
=∫[√(x-1)-1]dx/x
=∫√(x-1)dx/x -ln|x|
=2ln|√x+√(x-1)|-2√(1-x) -ln|x|+C

x=secu^2
∫√(x-1)dx/x=∫tanu*2secutanudu/secu^2)=2∫tanu^2cosudu=2∫sinu^2du/cosu=2∫du/cosu-2sinu
=2ln|secu+tanu|-2sinu+C=2ln|√x+√(x-1)|-2√(1-x)+C

解： 设根号(e^x-1) =t t^2 +1=e^x x=ln(t^2 +1) 代入得 ∫t dln(t^2 +1) =∫2t^2/(t^2 +1) dt =2*∫t^2/(t^2 +1) dt =2*∫(t^2 +1-1)/(t^2 +1) dt =2*∫[1 -1/(t^2 +1)] dt =2*[∫1 dt -∫1/(t^2 +1) dt =2*(t -arctant) +c(常数) =2*【(e^x-1) -arctan(e^x-1)】+c =2*【e^x -arctan(e^x-1)】+c(常数都归纳到c)