> 数学 >
求不定积分∫1/√(9-4x^2)dx=
人气:371 ℃ 时间:2020-02-05 17:52:02
解答
换元, 令u = 2x,
∫1/√(9-4x^2)dx = (1/2) ∫1/√(9- u^2) du
= (1/2) * arcsin(u/3) + C
= (1/2) * arcsin( 2x/3) + C
推荐
猜你喜欢
© 2024 79432.Com All Rights Reserved.
电脑版|手机版