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计算反常积分从负无穷到正无穷(4+9x²)分之1Dx

∫ (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)...

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