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