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Operations with negative sign

by rizardon
Tags: negative, operations, sign
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Mar18-14, 12:59 AM
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
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Mar18-14, 03:47 AM
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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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