damit "zerlegen" wir das mal

x^2+x+1 = 0
x\-{1,2} = -1/2 +/- j*sqrt(3)/2

1/(x^2+x+1)^2 = A/(x + 1/2 + j*sqrt(3)/2) + B/(x + 1/2 + j*sqrt(3)/2)^2 + C/(x + 1/2 - j*sqrt(3)/2) + D/(x + 1/2 - j*sqrt(3)/2)^2

1 = A*(x + 1/2 - j*sqrt(3)/2)*(x^2 + x + 1) + B*(x + 1/2 - j*sqrt(3)/2)^2 + C*(x + 1/2 + j*sqrt(3)/2)*(x^2 + x + 1) + D*(x + 1/2 + j*sqrt(3)/2)^2

1 = A*(x + 1/2 - j*sqrt(3)/2)*(x^2 + x + 1) + B*(x^2 + x - x*j*sqrt(3) - 1/2 - j*sqrt(3)/2) + C*(x + 1/2 + j*sqrt(3)/2)*(x^2 + x + 1) + D*(x^2 + x + x*j*sqrt(3) - 1/2 + j*sqrt(3)/2)

1 = A*(x^3 + 3x^2/2 - x^2*j*sqrt(3)/2 + 3x/2 - x*j*sqrt(3)/2 + 1/2 - j*sqrt(3)/2) + B*(x^2 + x - x*j*sqrt(3) - 1/2 - j*sqrt(3)/2) + C*(x^3 + 3x^2/2 + x^2*j*sqrt(3)/2 + 3x/2 + x*j*sqrt(3)/2 + 1/2 + j*sqrt(3)/2) + D*(x^2 + x + x*j*sqrt(3) - 1/2 + j*sqrt(3)/2)

I: A+C = 0

II: A*(3/2 - j*sqrt(3)/2) + B + C*(3/2 + j*sqrt(3)/2) + D = 0

III: A*(3/2 - j*sqrt(3)/2) + B*(1 - j*sqrt(3)) + C*(3/2 + j*sqrt(3)/2) + D*(1 + j*sqrt(3)) = 0

IV: A*(1/2 - j*sqrt(3)/2) + B*(-1/2 - j*sqrt(3)/2) + C*(1/2 + j*sqrt(3)/2) + D*(-1/2 + j*sqrt(3)/2) = 1

das ergibt dann

A = 2j/(3sqrt(3))
B = -1/3
C = -2j/(3sqrt(3))
D = -1/3

mit INT 1/(x+a)^2 dx = -1/(x+a) + C ergibt das

INT 1/(x^2+x+1)^2 dx = 2j/(3sqrt(3))*ln(x + 1/2 + j*sqrt(3)/2) + 1/(3x + 3/2 + j*3sqrt(3)/2) - 2j/(3sqrt(3))*ln(x + 1/2 - j*sqrt(3)/2) + 1/(3x + 3/2 - j*3sqrt(3)/2) + C
= 2j/(3sqrt(3))*[ ln(x + 1/2 + j*sqrt(3)/2) - ln(x + 1/2 - j*sqrt(3)/2) ] + [ 1/(3x + 3/2 + j*3sqrt(3)/2) + 1/(3x + 3/2 - j*3sqrt(3)/2) ] + C

es gilt  ln(a) - ln(b) = ln(a/b), daher:

INT 1/(x^2+x+1)^2 dx = 2j/(3sqrt(3))*ln((x + 1/2 + j*sqrt(3)/2) / (x + 1/2 - j*sqrt(3)/2)) + [ 1/(3x + 3/2 + j*3sqrt(3)/2) + 1/(3x + 3/2 - j*3sqrt(3)/2) ] + C

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nebenrechnung (1)

y = tan(x) = sin(x)/cos(x) = sinh(j*x)/(j*cosh(j*x)) = (e^(j*x) - e^(-j*x))/(j*(e^(j*x) + e^(-j*x))) = (e^(2j*x) - 1)/(j*(e^(2j*x) + 1))

y = (e^(2j*x) - 1)/(j*(e^(2j*x) + 1))
(j*(e^(2j*x) + 1))y = (e^(2j*x) - 1)
j*e^(2j*x)*y - e^(2j*x) = -1 - j*y
j*e^(2j*x)*y - e^(2j*x) = -1 - j*y
e^(2j*x)*(-1 + j*y) = -1 - j*y
e^(2j*x) = (-1 - j*y)/(-1 + j*y)
e^(2j*x) = (1 + j*y)/(1 - j*y)
2j*x = ln((1 + j*y)/(1 - j*y))
x = ln((1 + j*y)/(1 - j*y)) / (2j)
x = -j/2 * ln((1 + j*y)/(1 - j*y)) = arctan(y)
x = -j/2 * ln((j - y)/(j + y))
x = j/2 * ln((j + y)/(j - y))
x = j/2 * ln((y + j)/(y - j))

weiter, um auf unseren Term zu kommen

arctan(y) = j/2 * ln((y*sqrt(3)/2 + j*sqrt(3)/2)/(y*sqrt(3)/2 - j*sqrt(3)/2))
4arctan(y)/(3sqrt(3)) = 2j/(3sqrt(3)) * ln((y*sqrt(3)/2 + j*sqrt(3)/2)/(y*sqrt(3)/2 - j*sqrt(3)/2))

welchen wert für y muß ich nun einsetzen damit exakt der Term von oben entsteht

y*sqrt(3)/2 = x + 1/2
y = (2x+1)/sqrt(3)
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nebenrechnung (2)

1/(3x + 3/2 + j*3sqrt(3)/2) + 1/(3x + 3/2 - j*3sqrt(3)/2) =
1/3 * [ 1/(x + 1/2 + j*sqrt(3)/2) + 1/(x + 1/2 - j*sqrt(3)/2) ] =
1/3 * ( (x + 1/2 + j*sqrt(3)/2) + (x + 1/2 - j*sqrt(3)/2) ) / ( x^2 + x + 1 ) =
1/3 * ( 2x + 1 ) / ( x^2 + x + 1 )
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damit wird das zu

INT 1/(x^2+x+1)^2 dx = 4arctan((2x+1)/sqrt(3))/(3sqrt(3)) + 1/3 * ( 2x + 1 ) / ( x^2 + x + 1 ) + C

und voila, das ist genau das, was 'rauskommen soll;