Limit $\lim_{n\to \infty} n^2\int _0^1\frac{1}{(1+x^2)^n}$

Question

$\lim_{n\to \infty} n^2\int _0^1\frac{1}{(1+x^2)^n}=?.$

My attempt: I want to use Lebesgue Dominated convergence theorem to solve this, Because I see that for $\lim_{n\to \infty}\frac{n^2}{(1+x^2)^n}$ is zero on $ (0,1]$ so I split the integral into $[0,\epsilon]$ and $[\epsilon,1]$ and by Lebesgue Dominated convergence theorem the integral on $[\epsilon,1]$ is $0$ and since $\epsilon$ is arbitrary can I conclude the required limit to be $0.$

Answer 1

$\int_0^{\frac 1 {\sqrt n}} \frac {n^{2}} {(1+x^{2})^{n}}dx \geq \int_0^{\frac 1 {\sqrt n}} \frac {n^{2}} {(1+\frac 1 n)^{n}}dx \to \infty$.

Answer 2

Late answer since it is a duplicate somewhere else.

Note that for $n>1$

$$n^2\int_0^1\frac{1}{(1+x^2)^n}dx > n^2\int_0^1\frac{x}{(1+x^2)^n}dx$$ $$= \frac{n^2}{2} \int_0^1 \frac{d(1+x^2)}{(1+x^2)^n}$$ $$= \frac{n^2}{2(n-1)}\left(1-\frac{1}{2^{n-1}}\right)$$ $$> \frac{n}{4}$$