I encountered the following, deceptively simple looking integral - 

$$ \lim\_{n \to \infty} n^{\frac{3}{2}} \int\_0^1 \frac{x^2}{(1+x^2)^n} dx. $$

My goal was to hopefully somehow use the Lebesgue dominated convergence theorem; I've tried every elementary bound that I can think of to no avail. One approach suggested to me is to write 

$$ \int\_0^1 \frac{x^2}{(1+x^2)^{n+1}}dx = \int\_0^1 \frac{1}{(1+x^2)^n} - \int\_0^1 \frac{1}{(1+x^2)^{n+1}}dx, $$
then try and use an induction and partial fraction decomposition argument. Doing this, I didn't find anything useful.

When I tried plugging into wolfram/mathematica I got roughly the behavior (this could be wrong, so take it with a grain of salt) 
$$\int\_0^1 \frac{x^2}{(1+x^2)^n} dx = {}\_2 F\_1\left(\frac{1}{2}, n, \frac{3}{2}, -1\right). $$

Other calculations by trying to play with Gaussian bounds led to the result $ \frac{\sqrt{\pi}}{4} $ but this was done on a computer, not with an explicit method. There's also been some hints from mathematica that $$ \lim\_{n \to \infty} n^{\frac{3}{2}} \int\_0^1 \frac{x^2}{(1+x^2)^n} dx $$ is somehow related to $$ \sqrt{\pi}\frac{\Gamma\left(n-\frac{1}{2}\right)}{2\Gamma(n)}. $$

I believe that there is an elementary argument and that I just haven't found it. Any tips would be appreciated.