indefinite integration
回答 (2)
2011-05-29 7:20 am
✔ 最佳答案
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參考: miraco
2011-05-29 6:55 am
Sub u = 1 - √(1 - x), then du = dx/[2√(1 - x)]
dx = 2√(1 - x) du = 2(u - 1) du
So:
∫ dx/[1 - √(1 - x)]
= ∫ 2(u - 1)du/u
= ∫ (2 - 1/u) du
= 2u - ln |u| + C
= 2[1 - √(1 - x)] - ln |1 - √(1 - x)| + C
= -2√(1 - x) - ln |1 - √(1 - x)| + C since C is arbitrary
dx = 2√(1 - x) du = 2(u - 1) du
So:
∫ dx/[1 - √(1 - x)]
= ∫ 2(u - 1)du/u
= ∫ (2 - 1/u) du
= 2u - ln |u| + C
= 2[1 - √(1 - x)] - ln |1 - √(1 - x)| + C
= -2√(1 - x) - ln |1 - √(1 - x)| + C since C is arbitrary
參考: 原創答案
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