How fast does $a_n:=\int_{1/\sqrt{2}}^1 \frac{dx}{\left(\frac{1}{2}+x^2\right)^{n+1/2}}$ decay as $n \to \infty$? Let
\begin{equation}
a_n:=\int_{1/\sqrt{2}}^1 \frac{dx}{\left(\frac{1}{2}+x^2\right)^{n+1/2}}
\end{equation}
for $n \in \mathbb{N}$.
Then, $a_n$ is clearly a monotone decreasing sequence of positive numbers and by the Dominated Convergence Theorem, $a_n \to 0^+$ as $n \to \infty$.
However, I have some difficulty estimating how fast $a_n$ decays. For example, $\frac{a_n}{f(n)}=O(1)$ as $n \to \infty$ for some polynomial $f$?
Could anyone please provide insight into the decay rate of $a_n$?

If we perform the substitution $t = \sqrt {{\rm 2e}^s  - 1}$ in Svyatoslav's formula
$$
a_n  = 2^n \int_1^{\sqrt 2 } {\frac{{{\rm e}^{ - n\log (1 + t^2 )} }}{{\sqrt {1 + t^2 } }}{\rm d}t} ,
$$
we obtain
$$
a_n  = \frac{1}{{\sqrt 2 }}\int_0^{\log (3/2)} {{\rm e}^{ - ns} \frac{{{\rm d}s}}{{\sqrt {2 - {\rm e}^{ - s} } }}} .
$$
Now
$$
\frac{1}{{\sqrt {2 - {\rm e}^{ - s} } }} = \sum\limits_{k = 0}^\infty  {c_k \frac{{s^k }}{{k!}}} 
$$
for $|s|<\log 2$, where
$$\boxed{
c_k  = ( - 1)^k \sum\limits_{j = 0}^k {\frac{{(2j - 1)!!}}{{2^j }}S(k,j)} .}
$$
Here $j!!$ is the double factorial and $S(k,j)$ denotes the Stirling numbers of the second kind (cf. $\mathrm{A}352117$). Thus, by Watson's lemma,
$$\boxed{
a_n  \sim \frac{1}{{\sqrt 2 n}}\sum\limits_{k = 0}^\infty  {\frac{{c_k }}{{n^k }}}  = \frac{1}{{\sqrt 2 n}}\left( {1 - \frac{1}{{2n}} + \frac{5}{{4n^2 }} - \frac{{37}}{{8n^3 }} + \frac{{377}}{{16n^4 }} -  \ldots } \right)}
$$
as $n\to +\infty$. Note that, in fact,
$$
a_n  = \frac{1}{{\sqrt 2 n}}\sum\limits_{k = 0}^\infty  { P (k + 1,n\log (3/2))\frac{c_k}{{n^k }}} 
$$
where $P(a,z)$ is the normalised incomplete gamma function.
