求不定积分∫(√x)/[1-(√x)^(1/3)]dx
令√x=u³,则x=u⁶；dx=6u⁵du；
故原式=6∫[u⁸/(1-u)]du=-6∫[u⁸/(u-1)]du
=-6∫[u⁷+u⁶+u⁵+u⁴+u³+u²+u+1+1/(u-1)]du
=-6[u⁸/8+u⁷/7+u⁶/6+u⁵/5+u⁴/4+u³/3+u²/2+u+ln∣u-1∣]+C
=-6[(1/8)x^(4/3)+(1/6)x+(1/5)x^(5/6)+(1/4)x^(2/3)+(1/3)x^(1/2)+(1/2)x^(1/3)
+x^(1/6)+ln∣x^(1/6)-1∣+C