$X\sim N(1,1)$. Find $\operatorname{Var}(X^2)$ 
Let $X\sim N(1,1)$. Find $\operatorname{Var}(X^2)$.

I know since, $X-1\sim N(0,1)$, then $(X-1)^2\sim \chi^2_1$, hence $\operatorname{Var}(X-1)^2=2$. But don't understand how to find $\operatorname{Var}(X^2)$. Any help appreciated. Thanks.
 A: $\newcommand{\Var}{\operatorname{Var}} \newcommand{\E}{\mathbb{E}}$
You know $X=Z+1$ where $Z\sim N(0,1)$.
\begin{align}
\Var(X^2)&= \Var((Z+1)^2)\\
& = \Var(Z^2+2Z+1)\\
&=\Var(Z^2+2Z)\\
&=\E[Z^4+4Z^3+4Z^2]-(\E[Z^2]+2\E[Z])^2\\
&=\E[Z^4]+4\E[Z^3]+4\E[Z^2]-(1+0)^2\\
&\stackrel{(1)}{=}\Var(Z^2)+\E[Z^2]^2+0+4-1\\
&=2+1+4-1\\
&=6
\end{align}
Where in (1) we have used $\E[Z^{3}]=0$ (why?) and  $E[Z^4]=\Var(Z^2)+E[Z^2]^2$. Moreover I have seen that you already knew that $Z^2\sim \chi_1^2$ and therefore $\Var(Z^2)=2$.
A: $\newcommand{\e}{\operatorname E}$
$$
\operatorname{var}(X^2) = \e((X^2)^2) - (\e(X^2))^2.
 $$
 $$
 \e(X^2) = \operatorname{var}(X) + (\e(X))^2 = 1 + 1^2 = 2.
 $$
 Let $Z=X-1$ so that $Z\sim N(0,1).$ Then
\begin{align}
\e(X^4) & = \e((Z+1)^4) = \e(Z^4) + 4\e(Z^3) + 6\e(Z^2) + 4\e(Z) + 1. \\[10pt]
\e(Z^2) & = \operatorname{var}(Z) = 1. \\[10pt]
\e(Z^3) & = 0 \text{ by symmetry.} \\[10pt] {}
\end{align}
\begin{align}
\e(Z^4) & = \int_{-\infty}^\infty z^4 \varphi(z)\, dz = 2\int_0^\infty z^4\varphi(z)\, dz = 2\int_0^\infty z^4 \frac 1 {\sqrt{2\pi}} e^{-z^2/2} \, dz \\[10pt]
& = \sqrt{\frac 2 \pi} \int_0^\infty z^3 e^{-z^2/2} (z\,dz) = \sqrt{\frac 2 \pi} \int_0^\infty (2u)^{3/2} e^{-u} \, du \\[10pt]
& = \sqrt{\frac 2 \pi} \cdot 2^{3/2} \Gamma\left(\frac 5 2\right) = \frac 4 {\sqrt\pi} \cdot \frac 1 2 \cdot \frac 3 2 \cdot \Gamma\left( \frac 1 2 \right) \\[10pt]
& = 3.
\end{align}
