不定积分求解：∫(2x^2+2x+20)/[(x^2+2x+5)(x-1)]dx

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解决时间 2021-02-25 09:26

提问者网友：贪了杯
2021-02-24 20:07

最佳答案

五星知识达人网友：摆渡翁
2021-02-24 20:47

∫{ (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) + (B1x+B2)/(x^2+2x+5)2x-10≡ A(x^2+2x+5) +(B1x+B2)(x-1)x=18A=-8A=-1coef.of x^2A+B1=0B1=1coef.of constant5A-B2 =-10B2=5(2x-10)/[(x^2+2x+5)(x-1) ≡ -1/(x-1) + (x+5)/(x^2+2x+5)∫{ (2x^2+2x+20)/[(x^2+2x+5)(x-1)] }dx= 2∫ [1/(x-1) ]dx - ∫ (2x-10)/[(x^2+2x+5)(x-1)] dx=2∫ [1/(x-1) ]dx + ∫[1/(x-1)]dx - ∫ (x+5)/(x^2+2x+5) dx=3ln|x-1| - ∫ (x+5)/(x^2+2x+5) dx=3ln|x-1| - (1/2)∫ (2x+2)/(x^2+2x+5) dx - 4∫ [1/(x^2+2x+5) ]dx=3ln|x-1| -(1/2)ln|(x^2+2x+5)| -4∫ [1/(x^2+2x+5) ]dxconsiderx^2+2x+5 = (x+1)^2 +4letx+1 = 2tanydx = 2(secy)^2 dy∫ [1/(x^2+2x+5) ]dx=(1/2)∫ dy=y/2 + c'=(1/2)arctan[(x+1)/2] + C'∫{ (2x^2+2x+20)/[(x^2+2x+5)(x-1)] }dx=3ln|x-1| -(1/2)ln|(x^2+2x+5)| -4∫ [1/(x^2+2x+5) ]dx==3ln|x-1| -(1/2)ln|(x^2+2x+5)| -2arctan[(x+1)/2] + C

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1楼网友：由着我着迷
2021-02-24 21:39

好好学习下

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