（根号4减9x方）分之（1减x）的不定积分，要过程。。。。

Question

如题所述

Answer 1

∫ (1 - x)/√(4 - 9x²) dx
Let x = (2/3)sinz，dx = (2/3)cosz dz
√(4 - 9x²) = √(4 - 4sin²z) = 2cosz
=> ∫ (1 - (2/3)sinz)/(2cosz) * (2/3)cosz dz
= (1/3)∫ (1 - (2/3)sinz) dz
= (1/3)∫ dz - (2/9)∫ sinz dz
= (1/3)z - (2/9)(-cosz) + C
= (1/3)arcsin(3x/2) + (2/9)√(1 - (3x/2)²) + C
= (1/3)arcsin(3x/2) + (2/9)√(4 - 9x²)/2 + C
= (1/3)arcsin(3x/2) + (1/9)√(4 - 9x²) + C