令x^(1/6)=u,则x=u⁶,dx=6u⁵du,√x=u³,x^(1/3)=u²
原式=∫ [u³/(u³-u²)]6u⁵du
=6∫ u⁶/(u-1) du
=6∫ (u⁶-1+1)/(u-1) du
=6∫ (u⁶-1)/(u-1) du + 6∫ 1/(u-1) du
=6∫ (u⁵+u⁴+u³+u²+u+1) du + 6∫ 1/(u-1) du
=u⁶+(6/5)u⁵+(3/2)u⁴+2u³+3u²+6u+6ln|u-1|+C
=x + (6/5)x^(5/6) + (3/2)x^(2/3) + 2√x + 3x^(1/3) + 6x^(1/6) + 6ln|x^(1/6)-1| + C
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