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 $a_n$?

Here is a more elementary answer to close the debate. By the change of variable $t = n\,(x^2-1/2)$ (which gives $x^2 + 1/2 = 1+t/n$) one gets
$$
\int_{\frac{1}{\sqrt 2}}^1 \frac{\mathrm d x}{\left(\tfrac{1}{2}+x^2\right)^{n+1/2}} = \frac{1}{2n}\int_0^\frac{n}{2} \frac{\mathrm d t}{\left(1+\frac{t}{n}\right)^{n+1/2}\left(\frac{1}{2}+\frac{t}{n}\right)^{1/2}} = \frac{I_n}{2n}
$$
where $I_n = \int_0^\infty f_n$ with
$$
f_n(t) = \frac{\mathbb{1}_{[0,n]}(t)}{\left(1+\frac{t}{n}\right)^n\left(1+\frac{t}{n}\right)^{1/2}\left(\frac{1}{2}+\frac{t}{n}\right)^{1/2}} \underset{n\to\infty}\longrightarrow \frac{1}{e^t \left(\tfrac{1}{2}\right)^{1/2}}
$$
since $(1+t/n)^n \to e^t$. Hence, by dominated convergence,
$$
I_n = \int_0^\infty f_n \underset{n\to\infty}\longrightarrow\int_0^\infty \sqrt{2} \,e^{-t}\,\mathrm d t = \sqrt 2
$$
that is $I_n = \sqrt{2} + o(1)$. Thus,
$$
\int_{1/\sqrt 2}^1 \frac{\mathrm d x}{(\tfrac{1}{2}+x^2)^{n+1/2}} = \frac{1}{\sqrt 2}\frac{1}{n} + o\left(\frac{1}{n}\right).
$$
In particular, this agrees with the coefficient given by Svyatoslav.
