How to show $ \int_{-\infty}^{\infty} \frac{e^{-(x+1)^2}}{1+e^{-x}}\mathrm{d}x = \frac{\left(2\sqrt[4]{e} -1 \right)\sqrt{\pi}}{2e}$? I was recently looking at this post where the following formula is shown:
$$
\int_{-\infty}^{\infty} \frac{E(x)}{1+\mathcal{E}(x)^{O(x)}}\mathrm{d}x= \int_0^{\infty} E(x) \mathrm{d}x
$$
where $E(x), \mathcal{E}(x)$ are even functions and $O(x)$ is an odd function. One nice application of this formula would be the integral
$$
\int_{-\infty}^{\infty} \frac{e^{-x^2}}{1+e^{-x}}\mathrm{d}x = \frac{\sqrt{\pi}}{2}
$$
where the problem reduces to the evaluation of the Gaussian integral. I then wondered what would happen if I made slight alterations to the above integral, like changing $x^2\to (x+1)^2$. WA evaluates said integral as:

$$
 \int_{-\infty}^{\infty} \frac{e^{-(x+1)^2}}{1+e^{-x}}\mathrm{d}x = \frac{\left(2\sqrt[4]{e} -1 \right)\sqrt{\pi}}{2e}
$$


The even/odd formula can't be applied since the $+1$ makes the function not even anymore. Recalling that $\int^\infty_{-\infty} e^{-(ax^2 + bx+c)}\mathrm{d}x=\sqrt{\frac{\pi}{a}}e^{\frac{b^2}{4a}-c}
$ I attempted to evaluate the integral using geometric series
$$
\int_{-\infty}^{\infty} \frac{e^{-(x+1)^2}}{1+e^{-x}}\mathrm{d}x =  \sum_{n\ge 0}(-1)^n \int_{-\infty}^{\infty}e^{-(x^2+(n+2)x+1)}\, \mathrm{d}x = \frac{\sqrt{\pi}}{e}  \sum_{n\ge 0}(-1)^n e^{\frac{(n+2)^2}{4}}
$$
but the resulting series is divergent, so this method won't work. Does anyone have any ideas on how to evaluate this integral? Thank you!

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Hereafter, $\ds{\on{H}}$ is the $\ds{Heaviside\ Step\ Function}$.
\begin{align}
& \bbox[5px,#ffd]{\int_{-\infty}^{\infty}{\expo{-\pars{x + 1}^{\,\,\,2}} \over 1 + \expo{-x}}\,\dd x} =
\int_{-\infty}^{\infty}\expo{-\pars{x + 1}^{\,\,\,2}}
\bracks{\on{H}\pars{x} - {\on{sgn}\pars{x} \over
\expo{\verts{x}} + 1}}\dd x
\\[5mm] = & \
\int_{0}^{\infty}\bracks{\expo{-\pars{x + 1}^{\,\,\,2}} -
{\expo{-\pars{x + 1}^{\,\,\,2}} -
\expo{-\pars{-x + 1}^{\,\,\,2}}\over \expo{x} + 1}}\dd x
\\[5mm] = & \
\int_{0}^{\infty}\expo{-x^{2}\ -\ 1}
\pars{-1 + \expo{x} + \expo{-x}}\dd x =
{1 \over 2\expo{}}\int_{-\infty}^{\infty}\expo{-x^{2}}
\pars{-1 + \expo{x} + \expo{-x}}\dd x
\\[5mm] = & \
{1 \over 2\expo{}}\bracks{-\int_{-\infty}^{\infty}\expo{-x^{2}}
\dd x + \int_{-\infty}^{\infty}\expo{-\pars{x - 1/2}^{\,\,\,2}\,\, + 1/4}
\,\,\dd x \int_{-\infty}^{\infty}\expo{-\pars{x + 1/2}^{\,\,\,2}\,\, + 1/4}\,\,
\dd x}
\\[5mm] = & \
{1 \over 2\expo{}}\pars{-1 + \expo{1/4} + \expo{1/4}}
\int_{-\infty}^{\infty}\expo{-x^{2}}\,\,\dd x =
\bbx{{2\expo{1/4} - 1 \over 2\expo{}}\root{\pi}} \approx 0.5112
\end{align}
