You can put this solution on YOUR website! solve into partial fractions
(1/(x^3+1))
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= 1/[(x+1)(x^2-x+1)]
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= A/(x+1) + (Bx+C)/(x^2-x+1)
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Ax^2 -Ax + A + Bx^2+Bx + Cx + C = 1
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Equations:
A + B = 0
-A+B+C = 0
A + C = 1
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Solve this system of equations to get:
A = 1/3
B = -1/3
C = 2/3
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1/(x^3+1) = 1/(3(x+1)) + (-x+2)/[3(x^2-x+1)]
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Cheers,
Stan H.
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