Limit of this expression when n tends to infinity The limit is equal to: $$\lim_{n\to\infty} n^2 \int_0^1 \frac{1}{(1+x^2)^n} dx$$
P.S. What should be the better approach for such kinds of problems?

Assuming $n\geq 2$,
$$a_n=\int_{0}^{1}\frac{dx}{(x^2+1)^n} = \int_{0}^{\pi/4}\left(\cos\theta\right)^{2n-2}\,d\theta\leq \int_{0}^{\pi/4}e^{-(n-1)\theta^2}\,d\theta\leq \int_{0}^{+\infty}e^{-(n-1)\theta^2}\,d\theta=\frac{\sqrt{\pi}}{2\sqrt{n-1}} $$
since $\cos(x)\leq e^{-x^2/2}$ can be deduced by integrating $\tan(x)>x$. On the other hand the convexity of $\tan(x)$ implies that over $\left[0,\frac{\pi}{4}\right]$ we have $\log\cos(x)\leq -\frac{x^2}{3}$, so
$$ a_n \geq \int_{0}^{\pi/4}e^{-\frac{3}{2}(n-1)\theta^2}\,d\theta=\frac{\sqrt{\pi}}{\sqrt{6(n-1)}}-\int_{\pi/4}^{+\infty}e^{-\frac{3}{2}(n-1)\theta^2}\,d\theta $$
where
$$\int_{\pi/4}^{+\infty}e^{-\frac{3}{2}(n-1)\theta^2}\,d\theta\leq \int_{\pi/4}^{+\infty}e^{-\frac{3\pi}{8}(n-1)\theta}\,d\theta=o\left(\frac{1}{n}\right).$$
This gives that $\lim_{n\to +\infty}n^{\alpha}a_n$ equals $+\infty$ for any $\alpha>\frac{1}{2}$.
