Normalizing the Wavefunction

The probability density is

$dP(x) = A^{2} x^{2} e^{-2x^{2}} dx$

Therefore, we can do this:

$\int_{-\infty}^{+\infty} A^{2} x^{2} e^{-2x^{2}} dx = \left [-\frac{A^{2}}{2} \frac{\partial }{\partial \alpha} \int_{-\infty}^{+\infty} e^{-2\alpha x^{2}} dx \right ]_{\alpha = 1} = \frac{A^{2}}{4} \sqrt{\frac{\pi}{2}} = 1$

And then

$A = 2 \left (\frac{2}{\pi} \right )^{1/4}$