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November 28, 2014

November 28, 2014

Posted by **Sarah** on Friday, November 16, 2012 at 2:11am.

y=ln(1-x^2), 0<_x<_1/2

- CALCULUS -
**Steve**, Friday, November 16, 2012 at 11:12ams = ∫[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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