求不定积分∫ 1/√(x-x²)dx

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求不定积分∫ 1/√(x-x²)dx

∫ 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