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

发布网友 发布时间：2024-07-07 08:50

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热心网友 时间：2024-07-10 09:19

下图中的log代表ln

热心网友 时间：2024-07-10 09:19

∫{ (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)] dx

let
(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=1
8A=-8
A=-1

coef. of x^2
A+B1=0
B1=1

coef. of constant
5A-B2 =-10
B2=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) ]dx

consider
x^2+2x+5 = (x+1)^2 +4
let
x+1 = 2tany
dx = 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

热心网友 时间：2024-07-10 09:20

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