求不定积分∫(1-x)dx/√(9-4x²)
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∫(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/√(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
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