求不定积分∫1⼀√(9-4x^2)dx=

2025-04-13 16:28:20
推荐回答(2个)
回答1:

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

回答2:

凑微分,然后用arcsinx的积分公式。