Question

求不定积分∫(1-x)dx/√(9-4x²)？

Answer 1

∫(1-x)dx/√(9-4x²)
=∫1/√(9-4x²)dx-∫x/√(9-4x²)dx
=[∫1/√(9/4-x²)dx]/2-[∫1/√(9-4x²)dx²]/2
=∫1/√(9-4x²)dx+[∫1/√(9-4x²)d(9-4x²)]/8
=arcsin(2x/3)/2+√(9-4x²)/4+C
∫1/√(9-4x²)d(9-4x²)
=∫(9-4x²)^(-1/2)d(9-4x²)
=[1/(1-1/2)](9-4x²)^(1-1/2)
=2√(9-4x²)
你这里漏掉了2,1,∫(1-x)dx/√(9-4x²)
设 2x/3 = sint, 即 t = arcsin(2x/3)
∫(1-(3sint)/2)d(3sint/2)/√(9-9sin²t)
= ∫(1-(3sint)/2)d(3sint/2)/3cost
= (1/2) ∫(1-(3sint)/2)/cost * d(sint)
= (1/2) ∫(...,3,求不定积分∫(1-x)dx/√(9-4x²)

答案是(1/2)arcsin(2x/3)+√(9-4x²)/4+C
我算出来的是(1/2)arcsin(2x/3)+√(9-4x²)/8+C
就差一个数字
帮忙看看哪个错了?