令√(x+1)=t,则x=t^2-1
所以原式=∫(t-1)/(t+1)*2tdt
=2∫(t^2-t)/(t+1)dt
=2∫(t^2+t-2t-2+2)/(t+1)dt
=2∫tdt-4∫dt+4∫d(t+1)/(t+1)
=t^2-4t+4ln|t+1|+C
=x+1-4√(x+1)+4ln(√(x+1)+1)+C
=x-4√(x+1)+4ln(√(x+1)+1)+C
求不定积分 得解题思路及过程 ∫[√﹙x＋1﹚－1]／[√﹙x＋1﹚＋1]dx
令√(x+1)=t,则x=t^2-1
所以原式=∫(t-1)/(t+1)*2tdt
=2∫(t^2-t)/(t+1)dt
=2∫(t^2+t-2t-2+2)/(t+1)dt
=2∫tdt-4∫dt+4∫d(t+1)/(t+1)
=t^2-4t+4ln|t+1|+C
=x+1-4√(x+1)+4ln(√(x+1)+1)+C
=x-4√(x+1)+4ln(√(x+1)+1)+C