Find the exact length of the curve $y = \ln(1-x^2)$ Edit: Update with the full question for context

Find the exact length of the curve $y = \ln(1-x^2), 0 \leq x \leq \frac{1}{2}$

The integral below is what I got after finding the derivative $\frac{-2}{1-x^2}$ via the chain rule. 

Can someone give me a hint  on how to evaluate this integral with a range of $0 \leq x \leq \frac{1}{2}$?
$$ I = \int_{0}^{\frac{1}{2}} \sqrt{1+\bigg(\frac{-2x}{1-x^2}\bigg)^2}\mathrm dx$$
More specifically, how to deal with this fraction? That is what I am struggling the most with.

I tried to simplify it into $\frac{1}{1-2x^2 + x^4} = 1(1-2x^2+x^4)^{-1} = 1 - 2x^{-2} + x^{-4}$ which gives me
$$</s>I = \int_{0}^{\frac{1}{2}} \sqrt{2 - 2x^{-2} + x^{-4}} \mathrm dx$$
I looked at $u$-substitution, where $u = 2-2x^{-2} + x^{-4}$, but the $\mathrm du$ value didn't work out for me.
I am not entirely sure how trig substitution would work for a fraction//polynomial either. 

Partial fraction seems to be a good tool for your quesiton.
\begin{align}I &= \int_{0}^{\frac{1}{2}} \sqrt{1+\bigg(\frac{-2x}{1-x^2}\bigg)^2}\,\mathrm dx\\
&=\int_{0}^{\frac{1}{2}} \frac{\sqrt{(1-x^2)^2+4x^2}}{1-x^2}\mathrm \,dx\\
&=\int_0^\frac12 \frac{\sqrt{1+2x^2+x^4}}{1-x^2}\mathrm\, dx \\
&=\int_0^\frac12 \frac{1+x^2}{1-x^2}\mathrm\, dx \\
&=\int_0^\frac12 \frac{x^2-1+2}{1-x^2}\, dx \\
&= \int_0^\frac12 -1+\frac{2}{1-x^2}\, dx\\
&= \int_0^\frac12 -1 + \frac{1}{1-x} + \frac{1}{1+x} \, dx \\
&= -x - \ln (1-x)+\ln (1+x) |_{0}^{\frac12}\\
&= - \frac12 - \ln \frac12 + \ln \frac32 \\
&= \ln 3 - \frac12
\end{align}
