$$\int _0^1\frac{\ln (1+\sqrt{x})}{\sin(x)} \, dx\:$$
So the singularity point is at $0$, so we`ll use this test: $$\lim _{x\to 0}\frac{\frac{\ln(1+\sqrt{x})}{\sin(x)}}{\frac{1}{\sqrt{x}}}=\lim_{x\to \:0}\frac{\frac{\ln(1+\sqrt{x})}{\sin(x)}}{\frac{\sqrt{x}}{x}}=\lim_{x\to \:0}\frac{\ln(1+\sqrt{x})}{\sin(x)}\cdot \frac{x}{\sqrt{x}}=1$$ The last limit is calculated by L'Hopital's rule of $\lim_{x\to \:0}\frac{\ln(1+\sqrt{x})}{\sqrt{x}}$.
Therefore as $\int _0^1 \frac{1}{\sqrt{x}} \, dx<\infty$ exist so as $\lim_{x\to \:0}\frac{\ln(1+\sqrt{x})}{\sin(x)}<\infty $ Is it correct ?
The approach used by the OP is correct, but there is another approach that I thought would be instructive here. We note that
$$\sin x=x+O(x^3)$$
and
$$\log (1+x)=x+O(x^2)\implies \log (1+\sqrt{x})=\sqrt{x}+O(x)$$
Then, we have
$$\frac{\log (1+\sqrt{x})}{\sin x}=\frac{1}{\sqrt{x}}+O(\sqrt{x})$$
from which it is evident that the integral exists as an improper Riemann integral.