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Calculus

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Find the integral of
x^2 divided by the square root of (2x-x^2) dx

  • Calculus - ,

    x^2/(2x - x^2)
    = x/2 - 1

    so int(x^2/(2x - x^2) dx
    = int (x/2 - 1 ) dx

    = (1/4)x^2 - x + c

  • Calculus - ,

    eh? That's some fancy division there.
    I get

    2/(2-x) - 1

    so the integral is

    -2 ln(2-x) - x

    or some equivalent

  • Calculus - ,

    Oops. I missed that pesky square root.
    Let u = x-1 and we have

    integral (u+1)^2/√(1-u^2) du

    Now if u = sinθ
    du = cosθ dθ

    and we have integral of

    (1+sinθ)^2 dθ

    3/2 θ - 1/4 sin2θ - 2cosθ

    = 3/2 arcsin(u) - 1/2 u√(1-u^2) - 2√(1-u^2)

    = 3/2 arcsin(x-1) - 1/2 (x-1)√(2x-x^2) - 2√(2x-x^2)

    = 1/2 (3arcsin(x-1) - (x+3)√(2x-x^2))

  • Calculus - ,

    Ohhh my!!!

    Mea Culpa!!!

    That is embarrassing! What was I thinking??

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