x²/√x-x²的不定积分？
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√(x-x²) = √[1/4 - (x-1/2)²], 令 x - 1/2 = (1/2)sint, 则 dx = (1/2)costdt, 得
I = ∫[x²/√(x-x²)]dx = (1/4)∫[(1+sint)²dt = (1/4)∫[1+2sint+(sint)²]dt
= (1/4)∫[3/2+2sint-(1/2)cos2t]dt = (3/8)t - (1/2)cost - (1/8)sin2t + C
= (3/8)arcsin(2x-1) - √(x-x²) - (1/2)(2x-1)√(x-x²) + C
I = ∫[x²/√(x-x²)]dx = (1/4)∫[(1+sint)²dt = (1/4)∫[1+2sint+(sint)²]dt
= (1/4)∫[3/2+2sint-(1/2)cos2t]dt = (3/8)t - (1/2)cost - (1/8)sin2t + C
= (3/8)arcsin(2x-1) - √(x-x²) - (1/2)(2x-1)√(x-x²) + C
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