Limit evaluation with integral 
Evaluate the limit $$\lim_{n\to\infty} \int_0^1 n^2x(1-x^2)^n dx$$

My Proof: 
We may look at $n$ as a constant and evaluate the integral $\int_0^1 x(1-x^2)^ndx$ (I already moved out the $n^2$). From here we integrate by parts;
$$\int_0^1 x(1-x^2)^ndx = x \frac{(1-x^2)^{n+1}}{n+1}|_0^1 - \int_0^1 \frac{(1-x^2)^{n+1}}{n+1} dx = 0 - \frac{1}{n+1} \int_0^1 (1-x^2)^{n+1}dx = -\frac{1}{n+1} - ...$$
If I continue, something not very pretty happens. Is that even the preferable way of evaluating the limit?
 A: Hint: Let $t = x^2$ then $dt =2x dx$ and the integral becomes 
$$\frac{n^2}{2}\int_{0}^{1}(1 - t)^n dt = \frac{n^2}{2}\Bigg(-\frac{(1-t)^{n+1}}{n+1}\Bigg|_{0}^{1}\Bigg) $$
A: The easiest way to prove that the limit is $+\infty$, IMHO, is to prove that over the interval
$$I_n=\left[\frac{1}{\sqrt{n}},\frac{2}{\sqrt{n}}\right]$$
the function $f_n(x) = x(1-x^2)^n$ satisfies:
$$ f_n(x) \geq \frac{C}{\sqrt{n}} $$
for some absolute constant $C>0$. Since $f_n$ is non-negative over $(0,1)$, the previous inequality gives:
$$ \int_{0}^{1} n^2 f_n(x)\, dx \geq C n.$$
A: Express the integral in a form by which Laplace's method may be applied.  That is, rewrite the integral as
$$I(n) = n^2 \int_0^1 dx \, x \, e^{n \log{(1-x^2)}} $$
Most of the contribution from the integral takes place in the neighborhood about $x=0$, say, $x \in [0,\epsilon]$.  Thus we may Taylor expand the log about $x=0$ to get
$$I(n) = n^2 \int_0^{\epsilon} dx \, x \, e^{-n x^2 +O(n x^4)} $$
Note that we may assume $\epsilon = O\left ( n^{-1/2}\right )$.  Thus $n x^4 = O\left ( n^{-1}\right )$, and the exponential may be expanded.  Further, we may expend the integration limits to $[0,\infty)$ with exponentially small error.  Thus,
$$I(n) = n^2 \int_0^{\infty} dx \, x \, e^{-n x^2} \left [ 1+O\left ( \frac1{n}\right ) \right ] =\frac{n}{2}  +O\left ( 1\right ) $$
The limit is $\infty$.
