求不定积分 (1-√x-1)/1+三次根号下x-1

设 (x-1)^(1/6)=t,则 x=1+t^6,dx=6t^5；∫[1-√(x-1)]/[1+(x-1)^(¹/³)]dx=∫[(1-t³)/(1+t²)]*(6t^5)dt=6∫(1-t-t²+t³ +t^4 -t^6)+[(t-1)/(1+t²)] dt=6t-3t²-2t³+(3t^4 /2)-(6t^7 /7)+6∫[(t-1)/(1+t²)]dt=6t-3t²-2t³+(3t^4 /2)-(6t^7 /7)+3ln(1+t²)-6arctant+C；将 t 重新换为 (x-1)^(¹/6) 即可：=6(x-1)^(¹/6)-3(x-1)^(¹/³)-2√(x-1) +(3/2)(x-1)^(²/³)-(6/7)(x-1)(x-1)^(¹/6)+3ln[1+(x-1)^(¹/³)]-6arctan[(x-1)^(¹/6)]+C；