What is the integral of (x^3 - 1)^-1?

or 1 / (x^3 - 1)

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  • Anonymous
    1 decade ago
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    = 1/(x-1)(x+0.5+0.5sqrt(3)i)(x+0.5-0.5sqrt(3)i) = A/(x-1) + B/(x+0.5-0.5sqrt(3)i) + C/(x+0.5+0.5sqrt(3)i)

    lets find A,B,C

    multiply by x^3-1

    A(x^2+x+1) + B(x-1)(x+0.5+0.5 sqrt(3) i ) + C(x-1)(x+0.5-0.5sqrt(3) i ) = 1

    if x = 1 have 3A = 1 so A = 1/3

    if x = -0.5-0.5sqrt(3)i have C (-1.5 - 0.5sqrt(3)i)(sqrt(3)i) = 1 so

    C = (-1.5 sqrt(3)i + 1.5)^-1

    if x = -0.5+0.5sqrt(3)i, then

    B (-1.5 + 0.5sqrt(3)i)(-sqrt(3)i) = 1

    so B = (1.5sqrt(3)i + 1.5)^-1

    Integral of A(x-B)^-1 is Aln(x-B) + C

    Though, if you want it in R (real numbers, not complex) this wouldn't work for you. If so:

    assume 1/(x^3-1) = A/(x-1) + (Bx + C)/(x^2+x+1)

    A is 1/3 (already found)

    so multiply by (x^3-1)

    1 = 1/3(x^2+x+1) + (Bx + C)(x-1)

    -1/3x^2 - x/3 + 2/3 = Bx^2 + (C-B)x - C

    B = -1/3 , C = -2/3

    now we have our integral converted to

    1/3( (x-1)^-1 - (x + 2)/(x^2+x+1))

    (x-1)^-1 turns into ln(x-1)

    as to second, let p = x+1/2

    then x^2+x+1 turns into p^2+3/4

    so we have instead:

    - p/(p^2+3/4) - 1.5/(p^2+3/4) slightly better

    integral of - p/(p^2+3/4) is -0.5 ln (p^2+3/4)

    integral of - 1.5/(p^2+3/4) is -1.5 sqrt(3/4) atan (p / sqrt(3/4))

    p is x+1/2

    so

    1/3(ln(x-1) - 0.5 ln ((x+0.5)^2+3/4) -1.5 sqrt(3/4) atan ((x+0.5) / sqrt(3/4)) ) + const

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