设t=根号(x+1)x=t^2-1dx=2tdt∫dx/[x*根号下(1+x)]=∫2tdt/t(t^2-1)=2∫dt/(t+1)(t-1)=∫[1/(t-1)-1/(t+1)]dt=ln|t-1|-ln|t+1|+C=ln|(t-1)/(t+1)|+C再把t用根号(x+1)带回
分部积分
