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 $a_n$?

We can also try to find the asymptotics of the integral directly.
Making the substitution $x=\frac{t}{\sqrt2}$
$$a_n=\int_{1/\sqrt{2}}^1 \frac{dx}{\left(\frac{1}{2}+x^2\right)^{n+1/2}}=2^n\int_1^\sqrt2\frac{dt}{(1+t^2)^{n+1/2}}=2^n\int_1^\sqrt2\frac{e^{-n\ln(1+t^2)}}{\sqrt{1+t^2}}dt$$
Making one more substitution $t=1+x$
$$a_n=\frac{1}{\sqrt2}\int_0^{\sqrt{2}-1} \frac{e^{-n\ln(1+x+x^2/2)}}{\sqrt{1+x+x^2/2}}dx$$
It is straightforward to see that the function $f(x)=\ln(1+x+x^2/2)$ does not have an extremum on the interval $[0;\sqrt2-1]$, but has the minimal value at $x=0$. Decomposing the integrand near this point
$$\ln(1+x+x^2/2)=f(0)+f'(0)\,x+f''(0)\frac{x^2}{2}+...=x+O(x^3)$$
$$\frac{1}{\sqrt{1+x+x^2/2}}=1-\frac{1}{2}\,x+O(x^2)$$
$$a_n=\frac{1}{\sqrt2}\int_0^{\sqrt2-1}e^{-nx}\Big(1-\frac{x}{2}+O(x^2)\Big)dx$$
With the accuracy up to exponentially small corrections we can extend the interval of integration till $\infty$. Performing integration,
$$\boxed{\,\,a_n=\frac{1}{\sqrt2}\frac{1}{n}\Big(1-\frac{1}{2n}+O\big(1/n^2\big)\Big)\,\,}$$
Numerical check confirms the asymptotics:
$$n=10\quad a_n=0.0667968\quad \text{approximation}=\frac{1}{10\sqrt2}(1-1/20)=0.067175$$
$$n=30\quad a_n=0.0232066\quad \text{approximation}=\frac{1}{30\sqrt2}(1-1/60)=0.023177$$
$$n=100\quad a_n=0.0070366\quad \text{approximation}=\frac{1}{100\sqrt2}(1-1/200)= 0.00703571$$
$$n=1000\quad a_n=0.000706754\quad \text{approximation}=\frac{1}{1000\sqrt2}(1-1/2000)= 0.000706753$$
