Solution
= Solution
Assume $h$ is continuously <differentiable> on the interval concerned. For fixed $x>0$, take $u=h(t)$ and $dv=(x-t)^{-1/2}dt$, so $v=-2\sqrt{x-t}$. <Integration by parts>, interpreted first with an endpoint cutoff and then by a <limit>, gives
$$
g(x)=\left[-2h(t)\sqrt{x-t}\right]_{t=0}^{t=x}+2\int_0^xh'(t)\sqrt{x-t}\,dt=2h(0)\sqrt x+2\int_0^xh'(t)\sqrt{x-t}\,dt.
$$
The upper endpoint term vanishes, and the lower endpoint supplies the explicit square-root term.