I struggle with finalizing my argument and making sure all my steps are correct.
I realize it might be good to look at the integral this way: $\iint\limits_{\mathbb{R^2}} |\langle \begin{pmatrix} 1 \\ t\end{pmatrix} , \vec{x}\rangle| e^{-\langle \frac{1}{2}I\vec{x}, \vec{x}\rangle}dx$.
Then, use substitution in a way that $x' = A^{-1} x, A^{-1} = \begin{pmatrix} 1 & t \\ -t & 1\end{pmatrix}$, for the sake of simplifying the dot product. (Resulting with $A = \frac{1}{1+t^2} \begin{pmatrix} 1 & -t \\ t & 1\end{pmatrix}$)
In that way, I can get: $\iint\limits_{\mathbb{R^2}} |x_1| e^{-\langle \frac{1}{2}IA\vec{x}, A\vec{x}\rangle}|det{A}|dx_1dx_2$, which due to symmetry of 1st-4th quarters and 2nd-3rd quarters, results with
2$\iint\limits_{\mathbb{R^2\ right\ plane}} x_1 e^{-\langle A^t\frac{1}{2}IA\vec{x}, \vec{x}\rangle}dx_1dx_2$.
I am then stuck with making sure of my steps, as it looks like an integration by part might help, but it is abit tricky to me, and it looks like I want to take advantage of the fact that $\iint\limits_{\mathbb{R^n}} e^{-\langle B\vec{x}, \vec{x}\rangle}dx = \frac{\pi^{\frac{n}{2}}}{\sqrt{|detB|}}$ for positive definite matrices.
The solution should result with $\sqrt{8\pi(1+t^2)}$.
Thanks in advance to any helpers!
Following your approach we have that: $$\begin{cases} u=x+ty\\ v=-tx+y \end{cases} \implies \begin{cases}x=\frac{u-tv}{1+t^2}\\ y=\frac{tu+v}{1+t^2} \end{cases}.$$ Therefore $$\begin{align} \iint_{\mathbb{R}^2}{|x+ty|e^{-\frac{x^2+y^2}{2}}}dxdy &=\iint_{\mathbb{R}^2}{|u|e^{-\frac{(u-tv)^2+(tu+v)^2}{2(1+t^2)^2}}}\frac{dudv}{1+t^2} \\&=\frac{1}{1+t^2}\iint_{\mathbb{R}^2}{|u|e^{-\frac{u^2+v^2}{2(1+t^2)}}}dudv\\ &=\frac{1}{1+t^2}\int_{0}^{\infty}\int_{0}^{2\pi}{r|\cos(t)|e^{-\frac{r^2}{2(1+t^2)}}}rdrdt\\ &=\frac{4}{1+t^2}\int_{0}^{\infty}{r^2e^{-\frac{r^2}{2(1+t^2)}}}dr\\ &=4\sqrt{1+t^2}\int_{0}^{\infty}s^2e^{-s^2/2}ds\\ &=4\sqrt{1+t^2}\sqrt{\frac{\pi}{2}}=\sqrt{8\pi(1+t^2)} \end{align}$$ where $s=\frac{r}{\sqrt{1+t^2}}$ and we applied Integrate $x^2 e^{-x^2/2}$ .