∫t*(1+t)^1/2dt=∫[(1+t)-1](1+t)^1/2dt=∫[(1+t)^(3/2)-(1+t)^1/2]d(1+t)=(2/5)(t+1)^(5/2)-(2/3)(t+1)^(3/2)+C t*(1+t)^1/2的原函数是 (2/5)(t+1)^(5/2)-(2/3)(t+1)^(3/2)+C