求不定积分(2+x-x^2)^1⼀2dx

2024-07-27 17:45:51
推荐回答(2个)
回答1:

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/桐橘2)sin2t } + C
= (9/8) arcsin(2x-1)/3 + (1/2) (x-1/败轮拍2) √(2+x﹣x²察羡) + C

回答2:

积分:[(9/4)-(x-1/2)^2]^(1/2) dx
=sin^(-1)((x-1/伏皮春指2)/(3/2))
=sin^(-1)((2x-1)/缺森差3)