1/((1+x)(1-x^4))的不定积分?
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1-x^4 = (1-x^2)(1+x^2) =(1-x)(1+x)(1+x^2)
1/[(1+x)(1-x^4)] = 1/[(1-x)(1+x)^2.(1+x^2)]
let
1/[(1+x)(1-x^4)]≡ A/(1-x) +B/(1+x) +C/(1+x)^2 + (Dx+E)/(1+x^2)
=>
1≡ A(1+x)^2.(1+x^2) +B(1-x)(1+x)(1+x^2) +C(1-x)(1+x^2) + (Dx+E)(1-x)(1+x)^2
x=1, => A=1/8
x=-1, => C=1/4
x=i
1=(Di+E)(1-i)(1+i)^2
=2(Di+E)(1+i)
= (-2D+2E) +(2E +2D)i
=>
-2D+2E =1 (1)
2E +2D =0 (2)
(1)+(2)
4E=1
E=1/4
from (2)
2E +2D =0
D= -1/4
coef. of x^4
A-B-D =0
1/8-B+1/4 =0
B= 3/8
ie
1/[(1+x)(1-x^4)]
≡(1/8)[1/(1-x)] +(3/8)[1/(1+x)] +(1/4)[1/(1+x)^2] +(1/4) (-x+1)/(1+x^2)
∫ dx/[(1+x)(1-x^4)]
=∫ {(1/8)[1/(1-x)] +(3/8)[1/(1+x)] +(1/4)[1/(1+x)^2] +(1/4) (-x+1)/(1+x^2)} dx
=-(1/8)ln|1-x| +(3/8)ln|1+x| -(1/4)[1/(1+x)] +(1/4)∫(-x+1)/(1+x^2) dx
=-(1/8)ln|1-x| +(3/8)ln|1+x| -(1/4)[1/(1+x)] -(1/8)ln|1+x^2| +arctanx +C
1/[(1+x)(1-x^4)] = 1/[(1-x)(1+x)^2.(1+x^2)]
let
1/[(1+x)(1-x^4)]≡ A/(1-x) +B/(1+x) +C/(1+x)^2 + (Dx+E)/(1+x^2)
=>
1≡ A(1+x)^2.(1+x^2) +B(1-x)(1+x)(1+x^2) +C(1-x)(1+x^2) + (Dx+E)(1-x)(1+x)^2
x=1, => A=1/8
x=-1, => C=1/4
x=i
1=(Di+E)(1-i)(1+i)^2
=2(Di+E)(1+i)
= (-2D+2E) +(2E +2D)i
=>
-2D+2E =1 (1)
2E +2D =0 (2)
(1)+(2)
4E=1
E=1/4
from (2)
2E +2D =0
D= -1/4
coef. of x^4
A-B-D =0
1/8-B+1/4 =0
B= 3/8
ie
1/[(1+x)(1-x^4)]
≡(1/8)[1/(1-x)] +(3/8)[1/(1+x)] +(1/4)[1/(1+x)^2] +(1/4) (-x+1)/(1+x^2)
∫ dx/[(1+x)(1-x^4)]
=∫ {(1/8)[1/(1-x)] +(3/8)[1/(1+x)] +(1/4)[1/(1+x)^2] +(1/4) (-x+1)/(1+x^2)} dx
=-(1/8)ln|1-x| +(3/8)ln|1+x| -(1/4)[1/(1+x)] +(1/4)∫(-x+1)/(1+x^2) dx
=-(1/8)ln|1-x| +(3/8)ln|1+x| -(1/4)[1/(1+x)] -(1/8)ln|1+x^2| +arctanx +C
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