求1/(x^2)*√(x^2-9)dx的不定积分，下面有图，在线等，谢谢
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∫ √(9 - x²) dx
--> x = 3 siny，dx = 3 cosy dy -->
= ∫ √(9 - 9sin²y) 3cosy dy
= ∫ 3cosy 3cosy dy
= 9∫ cos²y dy
= (9/2)∫ (1 + cos2y) dy
= (9/2)[y + (1/2)sin2y] + C
= (9/2)y + (9/2)sinycosy + C
= (9/2)arcsin(x/3) + (9/2)(x/3)[√(9 - x²)/3] + C
= (9/2)arcsin(x/3) + (1/2)x√(9 - x²) + C
--> x = 3 siny，dx = 3 cosy dy -->
= ∫ √(9 - 9sin²y) 3cosy dy
= ∫ 3cosy 3cosy dy
= 9∫ cos²y dy
= (9/2)∫ (1 + cos2y) dy
= (9/2)[y + (1/2)sin2y] + C
= (9/2)y + (9/2)sinycosy + C
= (9/2)arcsin(x/3) + (9/2)(x/3)[√(9 - x²)/3] + C
= (9/2)arcsin(x/3) + (1/2)x√(9 - x²) + C
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