(23)
令⁶√x=t,则x=t⁶,³√x=t²,√x=t³
∫[√x/(³√x²-√x)]dx
=∫[t³/(t⁴-t³)]d(t⁶)
=∫[t³·6t⁵/(t⁴-t³)]dt
=6∫[t⁵/(t-1)]dt
=6∫[(t⁵-1+1)/(t-1)]dt
=6∫[t⁴ +t³+t²+t+1 +1/(t-1)]dt
=6[(1/5)t⁵+(1/4)t⁴+(1/3)t³+(1/2)t²+t] +6ln|t-1| +C
=(1/10)(12⁶√x⁵+15³√x²+20√x+30³√x+60⁶√x) +6ln|⁶√x-1| +C