= Solution
Choose the <holomorphic square root> in the <complex upper half-plane> whose boundary value is positive on $0<x<1$. Its primitive has <derivative> $F_1'(z)=1/\sqrt{z(1-z^2)}$. The integral from the boundary basepoint zero is well-defined by a limiting path: the local singularity is only $z^{-1/2}$.
The three finite <prevertices> $-1,0,1$ have <derivative> exponent $-1/2$, hence image angle $\pi/2$. At infinity $F_1'(z)=O(z^{-3/2})$, so the local coordinate $1/z$ also gives image angle $\pi/2$. These are the four corners of a <Schwarz-Christoffel mapping> to a <rectangle>. To establish that it is a <square> rather than a general <rectangle>, put
$$
A=\int_0^1\frac{dx}{\sqrt{x(1-x^2)}}.
$$
The length along $(-1,0)$ is also $A$ by $x\mapsto-x$. The length along $(1,\infty)$ is
$$
\int_1^\infty\frac{dx}{\sqrt{x(x^2-1)}}=A,
$$
by $x=1/t$, and the fourth side has the same length by <symmetry>. With the chosen branch, the boundary <derivatives> on the intervals $(-\infty,-1),(-1,0),(0,1),(1,\infty)$ have phases $-1,-i,1,i$, respectively. Therefore
$$
F_1(-1)=iA,\quad F_1(0)=0,\quad F_1(1)=A,\quad F_1(\infty)=A+iA.
$$
The extended real boundary maps continuously and once around this <square>. Each side is traversed monotonically, and the four corner limits are finite. The <argument principle>, applied after small indentations around the boundary singularities, counts one inverse image of each interior value and zero of each exterior value. Since $F_1'$ never vanishes in the <complex upper half-plane>, those interior inverse images are simple. Thus
$$
\boxed{F_1:\mathbb H\longrightarrow\{u+iv:0<u<A,\ 0<v<A\}\text{ is conformal and bijective}.}
$$
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