admin百科知識 2022-02-13 10:44:33 求不定積分 (1-√x-1)1+三次根號下x-1求不定積分 (1-√x-1)/1 三次根號下x-1網友廻答：匿名網友設 (x-1)^(1/6)=t,則 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... db標簽 生活常識_百科知識_各類知識大全»求不定積分 (1-√x-1)1+三次根號下x-1
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