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CALCULUS

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Find the length of the curve:
y=ln(1-x^2), 0<_x<_1/2

  • CALCULUS -

    s = ∫[0,1/2] ds
    = ∫[0,1/2] √(1+y'^2) dx
    = ∫[0,1/2] √(1+(2x/(x^2-1))^2) dx
    = ∫[0,1/2] (x^2+1)/(x^2-1) dx
    now use partial fractions to get
    = ∫[0,1/2] 1 + 1/(x-1) - 1/(x+1) dx
    = (x + ln|x-1| - ln|x+1|) [0,1/2]
    = (1/2 + ln(1/2) - ln(3/2)) - (0+0-0)
    = 1/2 - ln3

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