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

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

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

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