导数(1-x)/((9-4x^2)^(1/2))为的原函数是？

这个是求积分了：
∫(1-x)/√(9-4x²)dx
=∫1/√(9-4x²)dx - ∫x/√(9-4x²)dx
=(1/3)∫1/√[1-(2x/3)²]dx - (1/2)∫/√(9-4x²)dx²
=(1/2)∫1/√[1-(2x/3)²]d(2x/3) - (1/8)∫/√(9-4x²)d(2x)²
=(1/2)arcsin(2x/3) - (1/4)√(9-4x²) + C
原函数是：
(1/2)arcsin(2x/3) - (1/4)√(9-4x²) + C