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求不定积分 (1-√x-1)/1+三次根号下x-1
人气:415 ℃ 时间:2020-01-31 19:42:20
解答
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...
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