原式=∫x/(1-3x)^(1/3)dx ( 设t=(1-3x)^(1/3) 则 x=(1/3)(1-t^3) ,dx=-t^2 dt)
=(-1/3)∫(t-t^4)dt
=(-1/6)t^2+(1/15)t^5+C
= (-1/6)(1-3x)^(2/3)+(1/15)(1-3x)^(5/3)+C