Solving Arc Length Integral for y=ln(1-x^2) - 0 to 1/2

hachi_roku
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Homework Statement



ok, the original prob is : find the length of the curve of y=ln(1-x^2) x between 0, 1/2.



Homework Equations





The Attempt at a Solution


ive made it this far: my integral is -1 + 2/1-x^2.....ok so i decompose the second part but in doing so i get a negative to make it -(x+1)(x-1) but i don't know what happens to that negative because the solution manual says the integral is ...-1+ 1/x+1 -1/x-1 dx i don't get the signs. please help!
 
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if we are woking with single variable functions, y=f(x) like here then the arc length from a to be of a curve is:


L=\int_a^b\sqrt{1+[f'(x)]^2}dx
 
yes...like i said I've already worked that part...im toward the end of the problem i just don't get the signs
 
hachi_roku said:
yes...like i said I've already worked that part...im toward the end of the problem i just don't get the signs

Well, since you have shown almost no work(step by step) it is hard to tell where you have missed, or what you are doing wrong.
 
got it...nvm..thanks
 
hachi_roku said:
got it...nvm..thanks

:cool:
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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