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不定积分求解:∫(2x^2+2x+20)/[(x^2+2x+5)(x-1)]dx
人气:453 ℃ 时间:2020-05-11 18:07:50
解答
∫{ (2x^2+2x+20)/[(x^2+2x+5)(x-1)] }dx= 2∫{ (x^2+2x+5)/[(x^2+2x+5)(x-1)] }dx - ∫ (2x-10)/[(x^2+2x+5)(x-1)] dx= 2∫ [1/(x-1) ]dx - ∫ (2x-10)/[(x^2+2x+5)(x-1)] dxlet(2x-10)/[(x^2+2x+5)(x-1) ≡ A/(x-1...
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