calculus

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what is the integral of (x^2-x+6)/(x^3-3x)?

the process involves partial fractions, and the answer is supposed to include ln and arctan... i just don't know how to get there.

  • calculus -

    x^2-x+6
    ----------
    x (x^2-3)
    well the bottom is a bore since I can not really factor it ((x-sqrt 3)(x+sqrt 3) is no fun) so I will say
    a/(x) + (bx+c)/(x^2-3)
    then
    a(x^2-3) +(bx+c)(x) = 1 x^2 - 1 x + 6
    a x^2 - 3a + b x^2 + c x = 1 x^2 - 1 x + 6
    a + b = 1
    c = -1
    -3 a = 6
    so
    a = -2
    b = 3
    c = -1
    and
    -2/x + (3x-1)/(x^2-3) is what we have to integrate
    this is
    -2 integral dx/x + 3 integral x dx/(x^2-3) - integral (dx/(x^2-3)

  • calculus -

    f(x) = (x^2-x+6)/(x^3-3x)

    Factor the numerator:

    x^3 - 3x = x(x^2 - 3) =

    x[x-sqrt(3)][x+sqrt(3)]

    And it follows that the function f(x) must be of the form:

    f(x) = A/x + B/(x-sqrt(3)) +
    C/(x+sqrt(3))


    To find A multiply both sides by x and take the limit x --->0:

    A = -2

    To find B multiply both sides by
    (x-sqrt(3) ) and take the limit
    x --->sqrt(3):

    B = 3/2 - 1/6 sqrt(3)

    To find C multiply both sides by
    (x+sqrt(3) ) and take the limit
    x ---> -sqrt(3):

    C = 3/2 + 1/6 sqrt(3)

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