Given that $X_1$ and $X_2$ follow a normal distribution $N(\mu, \Sigma)$ whose mean and covariance are given by:
$\mu = \begin{bmatrix} 2 \\ 1 \end{bmatrix} $, $\Sigma = \begin{bmatrix} 4, -4 \\ -4, 8 \\ \end{bmatrix} $
Find $\mathbb{E}[X_1 | (X_1 + X_2)^2]$
How would I proceed with the square term: $(X_1 + X_2)^2$ in the condition?
Can I assume that it is also normally distributed?
Hint
Denote $f_{X_1,X_2}$ the distribution of $(X_1,X_2)$
Set $Z=(X_1+X_2)^2$. The distribution of $Z$ is given by $$f_Z(z)=\frac{\mathrm d }{\mathrm d z}\left(\int_{\mathbb R}\int_{-\sqrt{z}-x_1}^{\sqrt {z}-x_1}f_{X_1,X_2}(x_1,x_2)\,\mathrm d x_2\,\mathrm d x_1\right)\boldsymbol 1_{x>0}.$$ I let you do the calculation... I hope it's you'll find a closed form.
The joint distribution of $Z$ and $X_1$ is given by $$f_{X_1,Z}(x_1,z)=\frac{1}{2z}\left(f_{X_1,X_2}(x_1,\sqrt z-x_1)+f(x_1,-\sqrt z-x_1)\right)\boldsymbol 1_{z>0}.$$
Then $$\mathbb E[X_1\mid Z=z]=\int_{\mathbb R}x_1\frac{f_{X_1,Z}(x_1,z)}{f_Z(z)}\,\mathrm d x_1.$$
I would be very surprised if you can get a nice closed form.