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We are given that $\frac{d}{dx}F(x)=\frac{1}{\sqrt{2x-x^{2}}}$. This means that $F(x)$ is the antiderivative of $\frac{1}{\sqrt{2x-x^{2}}}$.
We need to integrate $\frac{1}{\sqrt{2x-x^{2}}}$. First, complete the square in the denominator: $2x - x^2 = -(x^2 - 2x) = -(x^2 - 2x + 1 - 1) = -( (x-1)^2 - 1) = 1 - (x-1)^2$. So, we have $\frac{1}{\sqrt{1 - (x-1)^2}}$.
Now we integrate: $\int \frac{1}{\sqrt{1 - (x-1)^2}} dx$. Let $u = x-1$, so $du = dx$. Then the integral becomes: $\int \frac{1}{\sqrt{1 - u^2}} du = \arcsin(u) + C = \arcsin(x-1) + C$. Thus, $F(x) = \arcsin(x-1) + C$.
We are given that $F(1) = 0$. Plugging this into our expression for $F(x)$: $F(1) = \arcsin(1-1) + C = \arcsin(0) + C = 0 + C = 0$. Therefore, $C = 0$.
So, $F(x) = \arcsin(x-1)$.
Final Answer: $F(x) = \arcsin(x-1)$
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