We've just started discussing continuous distributions in my probability class and I've come across this interesting distribution that I'm unsure of how to integrate. Let $c$ be a constant and let $X$ be a random variable with distribution $f(x)=ce^{-x^2-x}$. Find $c$.
By definition, we want to find $c$ such that $\int_{-\infty}^\infty ce^{-x^2-x} dx=1 $. However I'm unsure of how to integrate the $e^{-x^2-x}$ term. I first split the product such that we have $$c\int_{-\infty}^\infty e^{-x^2}e^{-x} dx $$ and then tried integration by parts with $u=e^{-x^2}$ but did not prove fruitful. I then attempted using $u=e^{-x}$. We know that $\int e^{-x^2}=\frac{\sqrt{\pi}}{2} \text{erf}(x)$. If we do parts, we end up with something like \begin{align*} c\int_{-\infty}^\infty e^{-x^2}e^{-x} dx = e^{-x}\frac{\sqrt{\pi}}{2} \text{erf}(x)\bigg\vert_{-\infty}^{\infty} - \int_{-\infty}^\infty \frac{\sqrt{\pi}}{2}\text{erf}(x)\left(-e^{-x}\right) dx \end{align*} which doesn't really get me anywhere. I feel like there's some obvious approach that I'm missing. I evaluated the integral on wolfram alpha and $\int_{-\infty}^\infty e^{-x^2-x} dx= \sqrt[4]{e}\sqrt{\pi}$ and the presence of the pi term tells me that I should try converting to polar coordinates first but that wasn't fruitful either. Any help is appreciated.
