Answer
The arc length function for the curve with starting point $(0,1)$ is given by
$$
\begin{aligned}
s(x)&=\int_{0}^{x} \sqrt{1+\left[y^{\prime}(t)\right]^{2}} d t \\
&=\int_{0}^{x} \sqrt{\frac{2}{1+t}} d t \\
&=\sqrt{2}[2 \sqrt{1+t}]_{0}^{x} \\
&=2 \sqrt{2}(\sqrt{1+x}-1).
\end{aligned}
$$
Work Step by Step
$$
y=\sin ^{-1} x+\sqrt{1-x^{2}}
$$
$\Rightarrow$
$$
\begin{aligned}
y^{\prime} &=\frac{1}{\sqrt{1-x^{2}}}-\frac{x}{\sqrt{1-x^{2}}}\\
&= \frac{1-x}{\sqrt{1-x^{2}}}
\end{aligned}
$$
$\Rightarrow$
$$
\begin{aligned} 1+\left(y^{\prime}\right)^{2}&=1+\frac{(1-x)^{2}}{1-x^{2}} \\
&=\frac{1-x^{2}+1-2 x+x^{2}}{1-x^{2}} \\
&=\frac{2-2 x}{1-x^{2}} \\
&=\frac{2(1-x)}{(1+x)(1-x)} \\
&=\frac{2}{(1+x)}
\end{aligned}
$$
$\Rightarrow$
$$
\sqrt {1+\left(y^{\prime}\right)^{2}}=\sqrt {\frac{2}{(1+x)} }
$$
So, the arc length function for the curve with starting point $(0,1)$ is given by
$$
\begin{aligned}
s(x)&=\int_{0}^{x} \sqrt{1+\left[y^{\prime}(t)\right]^{2}} d t \\
&=\int_{0}^{x} \sqrt{\frac{2}{1+t}} d t \\
&=\sqrt{2}[2 \sqrt{1+t}]_{0}^{x} \\
&=2 \sqrt{2}(\sqrt{1+x}-1).
\end{aligned}
$$
