可以先把x+2提出来
1/[(x-1)^(1/3)*(x+2)^(2/3)]
=1/(x+2) * [(x+2)/(x-1)]^(1/3)
做换元,令t=[(x+2)/(x-1)]^(1/3)
则x=1 + 3/(t^3-1)
dx=9t^2dt/(t^3-1)^2
代入原式得
∫3 / [t(t^3-1)] dt
=∫ -3/t + 1/(t-1) + (2t+1)/(t^2+t+1) dt
=-3ln|t| + ln|t-1| + ln(t^2+t+1) + C
再把t=[(x+2)/(x-1)]^(1/3)代入即可