这个积分题怎么做

let
x^(1/3) = sinu
(1/3)x^(-2/3) dx = cosu du
dx = 3(sinu)^2 .cosu du
∫arcsin[x^(1/3)] dx
=x.arcsin[x^(1/3)] -∫ {x/√[1-x^(2/3)]} . (1/3)x^(-2/3) dx
=x.arcsin[x^(1/3)] - (1/3) ∫ x^(1/3) /√[1-x^(2/3)] dx
=x.arcsin[x^(1/3)] - (1/3) ∫ (sinu /cosu) [ 3(sinu)^2 .cosu du ]
=x.arcsin[x^(1/3)] - ∫ (sinu)^3 du
=x.arcsin[x^(1/3)] + ∫ (sinu)^2 dcosu
=x.arcsin[x^(1/3)] + ∫ [1-(cosu)^2] dcosu
=x.arcsin[x^(1/3)] + cosu -(1/3)(cosu)^3 +C
=x.arcsin[x^(1/3)] + √[1-x^(2/3)] -(1/3)[1-x^(2/3)]^(3/2) +C