Solve the integral $\frac 1 {\sqrt {2 \pi t}}\int_{-\infty}^{\infty} x^2 e^{-\frac {x^2} {2t}}dx$ To find the Variance of a Wiener Process, $Var[W(t)]$, I have to compute the integral
$$
Var[W(t)]=\dots=\frac 1 {\sqrt {2 \pi t}}\int_{-\infty}^{\infty} x^2 e^{-\frac {x^2} {2t}}dx=\dots=t.
$$
I've tried integration by parts to solve the integral but end up with
$$
\dots=\frac 1 {\sqrt {2 \pi t}} \left(0 - \int_{-\infty}^{\infty} -\frac {x} {t} \cdot e^{-\frac {x^2} {2t}}\cdot \frac {x^3} 3 dx\right),
$$
which is even worse and probably wrong.
Can anyone please help me compute the first integral and show how it becomes equal to $t$?
(I know that the Variance for a Wiener Process (Standard Brownian motion)is defined as $t$ but want to prove it with the integral above.)

The easiest way to solve this integral is probably set t=1/u, then the function inside the integral is the derivative of the Gaussian density with respect to u.  After swapping integration and partial derivative (this is physics, so we don't need to prove that this is allowed) the remaining work is trivial.
If all you need is the solution, it is in https://en.wikipedia.org/wiki/List_of_integrals_of_Gaussian_functions
