Originally Posted by

**mollymcf2009** I think I did this correctly, but Webassign does not like my answer. Can someone please check my work? Thanks!! Molly

Find the arc length for:

$\displaystyle y = ln(1-x^2)$ on $\displaystyle 0 \leq x \leq \frac{1}{8}$

$\displaystyle L = \int\limits^{\frac{1}{8}}_{0} \sqrt{1 + \frac{4x^2}{x^4 - 2x^2 + 1}}dx$

$\displaystyle = \int\limits^{\frac{1}{8}}_{0} \sqrt{\frac{x^4 + 2x^2 + 1}{x^4 - 2x^2 + 1}}$

$\displaystyle = \int\limits^{\frac{1}{8}}_{0} \sqrt{\frac{(x^2+1)(x^2-1)}{(x^2-1)(x^2-1)}}$

$\displaystyle = \int\limits^{\frac{1}{8}}_{0} \frac{(x^2+1)}{(x^2-1)}$

$\displaystyle

= \int\limits^{\frac{1}{8}}_{0} 1 + \int\limits^{\frac{1}{8}}_{0} \frac{2}{x^2-1}$

$\displaystyle

= \int\limits^{\frac{1}{8}}_{0} 1 - \frac{1}{x+1} + \frac{1}{x-1}$

$\displaystyle = \biggl[ x - ln |x+1| + ln|x-1| \biggr]^{\frac{1}{8}}_0$

* If it looks good up to here, I think I may be botching the evaluation:

$\displaystyle = \biggl[\frac{1}{8} - ln(\frac{1}{8} + 1) + ln(\frac{1}{8} - 1)\biggr]$

$\displaystyle = \frac{1}{8} - ln\frac{9}{8} + ln\frac{7}{8}$

$\displaystyle = \frac{1}{8} - ln\frac{63}{8}$

Am I combining those log correctly? I'm drawing a blank.

Thanks!! Molly