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求不定积分(2+x-x^2)^1/2dx
人气:255 ℃ 时间:2020-05-03 21:01:52
解答
令 x﹣1/2 = (3/2) sint,dx = (3/2) cost dt
| = ∫ (9/4) cos²t dt
= 9/8 ∫ (1+ cos2t) dt
= (9/8) { t + (1/2)sin2t } + C
= (9/8) arcsin(2x-1)/3 + (1/2) (x-1/2) √(2+x﹣x²) + C
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