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dx/(4x-x^2)求不定积分
人气:170 ℃ 时间:2020-03-24 02:11:31
解答
∫ dx/(4x - x^2)
= ∫ dx/[x(4 - x)]
= (1/4)∫ [(4 - x) + x]/[x/(4 - x)] dx
= (1/4)∫ [1/x + 1/(4 - x)] dx
= (1/4)[ln(x) - ln(4 - x)] + C
= (1/4)ln[x/(4 - x)] + C
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