Let be $$ g_n(x):=\int_0^{x}\frac{t^{2n+1}}{e^t-1}\frac{1}{\sqrt{1-(t/x)^2}}dt$$ where we can compute
$\displaystyle \lim_{x\to 0}g_n(x)=0\quad$ and $\displaystyle \quad \lim_{x\to \infty}g_n(x)=\int_0^{\infty}\frac{t^{2n+1}}{e^t-1}dt=\text{Li}_{2 n+2}(1) \Gamma (2 n+2)$
Thus, as $g_n(x)$ exists and is continous $\forall x\in(0,+\infty)$, we can afirma that $g_n(x)$ is bounded in $[0,\infty)$, that is
$$ g_n(x)\le M_n\quad \forall x\in[0,\infty)\quad\text{with $M_n$ not depending on $x$}$$
I wonder if there is any way to compute this constant $M_n$.
NOTE: Graphics for $n=5$
