设√(1+4x)=t,则x=(t²-1)/4,dx=tdt/2故 原式=∫[(t²-1)/4]*t*tdt/2=1/8∫(t^4-t²)dt=1/8(t^5/5-t³/3)+C (C是积分常数)=(1+4x)^(5/2)/40-(1+4x)^(3/2)/24+C.