How to integrate 2x^2(e^-x^2) from 0 to infinity?

Using integration by parts,

Int [2x.x (e^-x²)] dx = 2x (-0.5 e^-x²) - Int [- e^-x²] dx.

Now I suppose you can use the standard result (from the normal distribution formula) in order to obtain the latter formula (adjusting the exponent from -0.5x² to -x² by a suitable transformation) and so work your way towards the final answer?

∫ [0, infi] 2 x^2 e^(-x^2) dx

Let u=x^2 ---> x=u^(1/2)
du = 2x dx

∫ [0, infi] 2 x^2 e^(-x^2) dx =∫ [0, infi] x e^(-u) dx = ∫ [0, infi] u^(1/2) e^(-u) du
when x=0, u=0
when x-->infinity, u-->infinity

= ∫ [0, infi] u^(3/2 -1) e^(-u) du
= Gamma(3/2)
= (1/2) Gamma(1/2)
= (1/2) sqrt(pi)
= sqrt(pi)/2

Note:
Gamma(n) = (n-1) Gamma(n-1)
Gamma(3/2) = (1/2) gamma(1/2)
gamma(1/2) = sqrt(pi)

http://en.wikipedia.org/wiki/Gamma_function