这道积分题怎么解，要过程，正解必采纳，谢谢啦。

∫√x/[1- x^(1/3)] dx

let
u= x^(1/6)
du = (1/6) x^(-5/6) dx
dx = 6u^5. du

∫[u^3/(1-u^2)] [6u^5. du]
=6∫u^8/(1-u^2) du
=6∫ [ -u^6-u^4-u^2-1 + 1/(1-u^2) ] du
=6[-(1/7)u^7 -(1/5)u^5 - (1/3)u^3 -u + arcsinu ] + C
=6[ -(1/7)x^(7/6) -(1/5)x^(5/6) - (1/3)x^(3/6) -x^(1/6) + arcsin(x^(1/6)) ] + C

consider
u^8
= -u^6(1-u^2) + u^6
=-u^6(1-u^2) - u^4(1-u^2) + u^4
=-u^6(1-u^2) - u^4(1-u^2) + u^2.(1-u^2) + u^2
=-u^6(1-u^2) - u^4(1-u^2) + u^2.(1-u^2) -(1- u^2) +1

解：

令⁶√x=t，则x=t⁶，√x=t³，³√x=t²

∫[√x/(1-³√x)]dx

=∫[t³/(1-t²)]d(t⁶)

=∫[t³·6t⁵/(1-t²)]dt

=6∫[t⁸/(1-t²)]dt

=6∫[(t⁸-1+1)/(1-t²)]dt

=6∫[(t⁸-1)/(1-t²)]dt+6∫[1/(1-t²)]dt

=6∫[(t⁴+1)(t²+1)(t²-1)/(1-t²)]dt+3∫[1/(1-t)+ 1/(1+t)]dt

=-6∫(t⁴+1)(t²+1)dt+3∫[1/(1-t)]dt +3∫[1/(1+t)]dt

=-6∫(t⁶+t⁴+t²+1)dt-3∫[1/(1-t)]d(1-t) +3∫[1/(1+t)]d(1+t)

=-6[(1/7)t⁷+(1/5)t⁵+(1/3)t³+t]-3ln|1-t|+3ln|1+t|+C

=(30t⁷+42t⁵+70t³+210t)/35 +3ln(|(1+t)/(1-t)|) +C

=[30x^(7/6)+42x^(5/6)+70√x+210x^(1/6)]/35 +3ln(|[1+x^(1/6)]/[1-x^(1/6)]|) +C