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

[hachi_roku]

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!

 

[sutupidmath]

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:


$$L=\int_a^b\sqrt{1+[f'(x)]^2}dx$$

 

[hachi_roku]

yes...like i said I've already worked that part...im toward the end of the problem i just don't get the signs

 

[sutupidmath]

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.

 

[hachi_roku]

got it...nvm..thanks

 

[sutupidmath]

got it...nvm..thanks