These two methods require external/prior knowledge, which is highlighted below
METHOD 1
let # I = int_{-oo}^{oo} x^2 \ e^[(- x^2)/2] dx#
# = 2 int_{0}^{oo} x^2 \ e^[(- x^2)/2] dx# as its symmetrical
#t = x^2/2, dt = x dx = sqrt(2t) \ dx#
# I = 2 int_{0}^{oo} 2t \ e^[- t] 1/sqrt(2t) \ dt#
# I =2 sqrt2 int_{0}^{oo} sqrt(t) \ e^[- t] \ dt qquad triangle#
PRIOR KNOWLEDGE: #mathcal(L) { sqrt(t)}(s) = 1/(2s) sqrt(pi/s)#
and #triangle# is simply the Laplace transform of #sqrt(t)# with s = 1
So #I =2 sqrt2* 1/2 sqrt pi = sqrt (2 pi)#
METHOD 2
PRIOR KNOWLEDGE: #int_(-oo)^(oo) \ e^(-alpha x^2) = sqrt(pi/alpha)#
differentiate wrt #alpha#
#d/(d alpha) int_(-oo)^(oo) \ e^(-alpha x^2) = d/(d alpha) sqrt(pi/alpha)#
# int_(-oo)^(oo) -x^2 \ e^(-alpha x^2) = -1/2 sqrt(pi) alpha^(-3/2) #
# int_(-oo)^(oo) x^2 \ e^(-alpha x^2) = 1/2 sqrt(pi/ alpha^3)#
#alpha = 1/2#
# int_(-oo)^(oo) x^2 \ e^(-x^2/2) = 1/2 sqrt(pi/ (1/2)^3) = sqrt (2 pi)#
that's the same answer with two slightly different approaches.
