$\int_{0}^{\pi}\frac{ \sqrt{sin(3x)}}{x\left(e^{3\sqrt{x}}-1)\right)}dx$
I know
$\int_{0}^{\pi}\frac{ \sqrt{sin(3x)}}{x\left(e^{3\sqrt{x}}-1)\right)}dx = \lim_{c\to 0}\int_{c}^{\pi}\frac{ \sqrt{sin(3x)}}{x\left(e^{3\sqrt{x}}-1)\right)}dx$
I think it needs to be a comparison test but do not know how to proceed.
When $x\approx 0$ you have $$ \sqrt{\sin 3x}=\sqrt{3x}+o(\sqrt x),\quad x\to 0 $$ and $$ x(e^{3\sqrt x}-1)=3x\sqrt{x}+o(x^{3/2}),\quad x\to 0 $$ Therefore, the integrand is $$ f(x)=\frac 1{\sqrt{3}x}+... $$ and the integral is divergent.
You can also use “limit comparison” with $g(x)=1/x$.