令 1+(x+1)^(1/3) = t,则 x=(t-1)^3-1,dx=3(t-1)^2dt
I = ∫ dx/[1+(x+1)^(1/3)] = ∫ 3(t-1)^2dt/t
= 3 ∫ (t-2+1/t)dt = 3( t^2/2-2t+ln|t| ) + C
= (3/2) [1+(x+1)^(1/3)]^2 -6[1+(x+1)^(1/3)]+3ln|1+(x+1)^(1/3)| + C