Find differents methods Let $I_n=\int_0^1 \frac 1{(1+x^2)^n} dx$ 
With Laplace's method, i find easlly that $I_n\sim \frac 12\sqrt{\frac \pi n}$
I  would like to find  differents methods to prove it 
 A: We will prove that
\begin{equation*}
 \lim_{n\to \infty}\dfrac{I_n}{\frac{1}{2}\sqrt{\frac{\pi}{n}}} =1.
\end{equation*}
After the substitution $x=\frac{y}{\sqrt{n}}$ we have that
\begin{equation*}
 I_n = \dfrac{1}{\sqrt{n}}\int_{0}^{\sqrt{n}}\dfrac{1}{\left(1+\frac{y^2}{n}\right)^{n}}\, \mathrm{d}y .
\end{equation*}
Put
\begin{equation*}
f_n(y) =
 \begin{cases}
  \dfrac{1}{\left(1+\frac{y^2}{n}\right)^{n}}&\mbox{ if } 0<y<\sqrt{n}\\
  0&\mbox{ if } y\ge \sqrt{n}
 \end{cases}
\end{equation*}
It is well known that $\left(1+\frac{y^2}{n}\right)^{n}$ is increasing to $e^{y^2}$.
Consequently $0\le f_n(y) \le \dfrac{1}{1+y^2} \in L_1(\mathbb{R}_{+})$ and $\displaystyle \lim_{n\to \infty}f_n(y) = e^{-y^2}$ pointwise. According to Lebesgue's dominated convergence theorem we get that
\begin{gather*}
 \dfrac{I_n}{\frac{1}{2}\sqrt{\frac{\pi}{n}}} =\dfrac{2}{\sqrt{\pi}}\int_{0}^{\sqrt{n}}\dfrac{1}{\left(1+\frac{y^2}{n}\right)^{n}}\, \mathrm{d}y = \dfrac{2}{\sqrt{\pi}}\int_{0}^{\infty}f_n(y)\, \mathrm{d}y \to \\[2ex] \dfrac{2}{\sqrt{\pi}}\int_{0}^{\infty}e^{-y^2}\, \mathrm{d}y =1 , \mbox{ as } n\to \infty .
\end{gather*}
A: First set $x=\tan u$. Then your integral becomes
$$
I_n=\int_0^{\pi/4}\cos^{2n-2}(u)\,du.
$$
It would have been nicer to have the integral from $0$ to $\pi/2$. Since
$$
\Bigl|\int_{\pi/4}^{\pi/2}\cos^{2n-2}(u)\,du\Bigr|\leq\frac{\pi}{4}\Bigl(\frac{1}{\sqrt{2}}\Bigr)^{2n-2}=\frac{\pi}{2^{n+1}}
$$
this integral is exponentially small, and we can add it without changing the asymptotics (as long as we find that the result is not exponential small).
The integral from $0$ to $\pi/2$ can be calculated explicitly,
$$
\int_0^{\pi/2}\cos^{2n-2}(u)\,du=\frac{\sqrt{\pi}\,\Gamma(n-1/2)}{2\Gamma(n)}.
$$
The suggested result follows by using the Stirling approximation of $\Gamma$.
