Marginal distribution of $X$ given $f_{XY}(x,y)=ce^{-x^2-y^2-xy-x}$ I'm supposed to calculate the marginal distribution of $X$ given that the joint distribution is $ce^{-x^2-y^2-xy-x}$ for some $c>0$ and I know that that means $$f_X(x)=c\int_{-\infty}^\infty e^{-(x^2+y^2)-xy-x}dx$$ and I learned in class that you can use polar coordinates to simplify integrals similar to this but I am not really sure if that can be applied to this problem? Any tips on how to proceed would be greatly appreciated, thanks!

One way to integrate is to complete the square in the exponent, firstly for $y$ and then for $x$, to help you find $c$. We want to put the integrand into the form of a Standard Normal pdf so we also introduce a factor of $\frac{1}{2}$:
\begin{align}
1 &= c \int_{x=-\infty}^{\infty} \int_{y=-\infty}^{\infty} \exp\left[-\dfrac{1}{2}\left(\sqrt{2}y + \dfrac{x}{\sqrt{2}}\right)^2 -\dfrac{1}{2}\left(\sqrt{\dfrac{3}{2}}x + \sqrt{\dfrac{2}{3}}\right)^2 + \dfrac{1}{3} \right]dy\;dx. \\
\end{align}
Letting $u = \sqrt{2}y + \dfrac{x}{\sqrt{2}},$ and $v = \sqrt{\dfrac{3}{2}}x + \sqrt{\dfrac{2}{3}}$, we have $dy=\dfrac{du}{\sqrt{2}}$ and $dx=\sqrt{\dfrac{2}{3}}dv,$ so:
\begin{align}
1 &= \dfrac{2\pi e^{1/3}}{\sqrt{3}} c \int_{u=-\infty}^{\infty} \dfrac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}u^2} du \int_{v=-\infty}^{\infty} \dfrac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}v^2} dv \\
&= \dfrac{2\pi e^{1/3} c}{\sqrt{3}} \\
\therefore\quad c &= \dfrac{\sqrt{3}}{2\pi e^{1/3}}.
\end{align}
The same technique, but just integrating over $y$, allows you to then find the marginal pdf $f_X(x)$.
