# Operations with negative sign

by rizardon
Tags: negative, operations, sign
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 P: 20 While working on an integration problem I found that I will arrive at two different solutions depending on how I approach it. I'm finding the arc length of y=ln(1-x2) on the interval [0,0.5] The formula for finding the arc length is ∫sqrt[1+[f'(x)]2]dx So f'(x) = -2x / ( 1-x2 ) Here I first simplify this to 2x / ( x2 - 1 ) and squaring gives 4x2 / ( x2 -1 )2 Working from here I end up integrating from 0 to 0.5 ∫ [1 + 1/(x-1) - 1/(x+1)] dx = 0.5 - ln3 On the other hand if I leave f'(x) as it is without simplifying, when I squared f'(x) I get 4x2 / ( 1-x2 )2 and end up integrating from 0 to 0.5 ∫ [1 + 1/(1+x) + 1/(1-x)] dx = -0.5 - ln3 Should both have the same solution or is this simply a possible effect from squaring numbers? Thank you
 Sci Advisor HW Helper PF Gold P: 12,016 Note that when taking the square root of the perfect square in your denominator, you must use the ABSOLUTE value, |x^2-1| as your new denominator.

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