Question

不定积分的计算？

软件答案跟题目给的答案不一样，有没有简便方法

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Answer 1

I = ∫dx/√[x(1-x)] = ∫dx/√[x-x^2] = ∫dx/√[1/4-(x-1/2)^2]
令 x-1/2 = (1/2)sint， 则 dx = (1/2)costdt
I = ∫ (1/2)costdt/(1/2)cost] = ∫dt = t + C = arcsin(2x-1) + C
题目与答案不对应。