换元法.令t=√(x+1) 则x=t^2-1 dx=2tdt; ∫x√x+1dx=∫2t^2(t^2-1)dt =∫(2t^4-2t^2)dt =(2/5)t^5-(2/3)t^3+C 由t=√(x+1) =(2/5)(x+1)^(5/2)-(2/3)(x+1)^(3/2)+C
