Answer
The exact length of the curve
$$
y=\ln(1-x^{2}), \quad\quad 0 \leq x \leq \frac{1}{2}.
$$
is equal to $ (\ln 3-\frac{1}{2}). $
Work Step by Step
$$
y=\ln(1-x^{2}), \quad\quad 0 \leq x \leq \frac{1}{2}.
$$
We have
$$
y=\ln(1-x^{2}) \Rightarrow d y/ d x=\frac{1}{1-x^{2}}. (-2x)
$$
$$
\begin{aligned}
1+\left(\frac{d y}{d x}\right)^{2} &=1+\frac{4 x^{2}}{\left(1-x^{2}\right)^{2}} \\
&=\frac{1-2 x^{2}+x^{4}+4 x^{2}}{\left(1-x^{2}\right)^{2}}\\
&=\frac{1+2 x^{2}+x^{4}}{\left(1-x^{2}\right)^{2}}\\
&=\frac{\left(1+x^{2}\right)^{2}}{\left(1-x^{2}\right)^{2}}
\end{aligned}
$$
$
\Rightarrow
$
$$
\begin{aligned}
\sqrt{1+\left(\frac{d y}{d x}\right)^{2}}& =\sqrt{\left(\frac{1+x^{2}}{1-x^{2}}\right)^{2}} \\
&=\frac{1+x^{2}}{1-x^{2}}\\
& \quad\left[ \text{ by long division } \right]\\
&=-1+\frac{2}{1-x^{2}}\\
& \quad\left[ \text{ its partial fractions are } \right]\\
&=-1+\frac{1}{1+x}+\frac{1}{1-x}
\end{aligned}
$$
So the arc length is
$$
\begin{aligned}
L &=\int_{0}^{1 / 2}\left(-1+\frac{1}{1+x}+\frac{1}{1-x}\right) d x \\
&=[-x+\ln |1+x|-\ln |1-x|]_{0}^{1 / 2} \\
&=\left(-\frac{1}{2}+\ln \frac{3}{2}-\ln \frac{1}{2}\right)-0 \\
&=\ln 3-\frac{1}{2}
\end{aligned}
$$
Hence, the exact length of the curve
$$
y=\ln(1-x^{2}), \quad\quad 0 \leq x \leq \frac{1}{2}.
$$
is equal to $ (\ln 3-\frac{1}{2}). $
