由题意可得：∫[x/√(x+1)]dx=∫[(x+1-1)/√(x+1)]dx=∫√(x+1)dx-∫1/(√(x+1)dx=∫√(x+1)d(x+1)-∫1/(√(x+1)d(x+1)=2[(x+1)^(3/2)]/3-2√(x+1)+C又积分上限为3,下限为0...