= Solution
The additional <branch points> $\pm1/k$ lie on the two deleted rays because $0<k<1$. On the same <simply connected domain>
$$
D=\mathbb C\setminus\bigl((-\infty,-1]\cup[1,\infty)\bigr),
$$
choose both square roots to equal $1$ at zero. Their product has a single-valued holomorphic reciprocal on $D$, so its integral from zero defines the single-valued branch $H$ of the <elliptic integral of the first kind>.
An unrestricted contour may wind around any of $-1,1,-1/k,1/k$. Continuation around one such point flips one square root, so the value can change by <monodromy reflection at a square-root branch point>[square-root monodromy]. Thus $h$ is multivalued.
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