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求不定积分∫ 1/√(x-x²)dx
人气:499 ℃ 时间:2019-10-30 07:31:40
解答
∫ 1/√(x-x²)dx
=∫ 1/√(1/4-1/4+x-x²)dx
=∫ 1/√(1/4-(x-1/2)²)dx
=∫ 1/√(1/4-(x-1/2)²)d(x-1/2)
=arcsin(x-1/2)/(1/2)+c
=arcsin(2x-1)+c
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