I was observing this question by @Brightsun and conjecture $(1)$
$$\int_{0}^{\infty}e^{-\sqrt{x}}\ln{\left(1+{1\over \sqrt{x}}\right)}=2\gamma \tag1$$
An attempt
$x=u^2$ then $(1)$ becomes
$$2\int_{0}^{\infty}ue^{-u}\ln{\left(1+{1\over u}\right)}\tag2$$
$$2\sum_{n=0}^{\infty}{(-1)^n\over n!}\int_{0}^{\infty}u^{n+1}\ln{\left(1+{1\over u}\right)}\mathrm du\tag3$$
Changing $(2)$ by applying $e^x$, $(3)$ diverges.
How would one prove $(1)?$
By the change of variable $$ u=\sqrt{x},\quad x=u^2, \quad dx=2udu, $$ one has $$ \begin{align} \int_{0}^{\infty}e^{-\sqrt{x}}\ln{\left(1+{1\over \sqrt{x}}\right)}dx&=2\int_{0}^{\infty}u\:e^{-u}\ln{\left(1+{1\over u}\right)}du \\\\&=2\int_{0}^{\infty}u\:e^{-u}\ln{\left(1+u\right)}\:du-2\int_{0}^{\infty}u\:e^{-u}\ln{u}\:du. \tag1 \end{align} $$ By integrating by parts, the first integral on the right hand side of $(1)$ gives $$ \begin{align} &\int_{0}^{\infty}u\:e^{-u}\ln{\left(1+u\right)}\:du \\\\&=\left[\frac{e^{-u}}{-1}\cdot u\:\ln{\left(1+u\right)}\right]_{0}^{\infty}+\int_{0}^{\infty}e^{-u}\left(\ln{\left(1+u\right)}+\frac{u}{1+u}\right)du \\\\&=\int_{0}^{\infty}e^{-u}\left(\ln{\left(1+u\right)}+\frac{u+1-1}{1+u}\right)du \\\\&=\int_{0}^{\infty}e^{-u}\left(\color{red}{\ln{\left(1+u\right)}-\frac{1}{1+u}}\right)du+\int_{0}^{\infty}e^{-u}du \\\\&=\color{red}{0}+1 \quad (\text{integration by parts}) \end{align} $$ By integrating by parts, the second integral on the right hand side of $(1)$ gives $$ \begin{align} &\int_{0}^{\infty}u\:e^{-u}\ln u\:du \\\\&=\left[\frac{e^{-u}}{-1}\cdot u\:\ln u\right]_{0}^{\infty}+\int_{0}^{\infty}e^{-u}\left(\ln u+1\right)du \\\\&=\int_{0}^{\infty}e^{-u}\ln u\:du+1 \end{align} $$ yielding $$ \int_{0}^{\infty}e^{-\sqrt{x}}\ln{\left(1+{1\over \sqrt{x}}\right)}dx=-2\int_{0}^{\infty}e^{-u}\ln u\:du, $$ which is the announced result.
Note that $$\begin{eqnarray*} \frac{d}{du}\left[e^{-u}\left(u\ln u-(1+u)\ln(1+u)\right)\right]\hspace{-40mm}&&\\ &=&e^{-u}\left(\ln u-\ln(1+u)-u\ln u+(1+u)\ln(1+u)\right)\\ &=&e^{-u}\left(\ln u+u\ln(1+1/u)\right). \end{eqnarray*}$$ Hence $$\begin{eqnarray*} 2\int_0^\infty ue^{-u}\ln(1+1/u)\;du &=&2\left[e^{-u}\left(u\ln u-(1+u)\ln(1+u)\right)\right]_0^\infty\\ &&{}-2\int_0^\infty e^{-u}\ln u\;du\\ &=&2\gamma \end{eqnarray*}$$ by the first formula here.
Starting from $$2\int_0^\infty u e^{-u}\ln\left(1+\frac{1}{u}\right)du =2\int_0^\infty u e^{-u}\left(\ln(1+u)-\ln(u)\right)du$$ Let's find $\int_0^\infty u e^{-u}\ln(u)du$ first. We can introduce the variable $a$ and examine the integral: $$\int_0^\infty e^{-u} u^adu=\Gamma(a+1)$$ Now we will take the derivative by $a$ to get back to our original integral: $$\frac{\partial}{\partial a}\int_0^\infty e^{-u} u^adu =\int_0^\infty e^{-u}u^a\ln(u)du=\Gamma'(a+1)$$ Let $a=1$. We then have $$\int_0^\infty e^{-u}u\ln(u)du=\Gamma'(2)=1-\gamma$$ Now $$\int_0^\infty ue^{-u}\ln(1+u)du=\left[-e^{-u}u\ln(1+u)\right]_0^\infty+ \int_0^\infty e^{-u}\left(\ln(1+u)+\frac{u}{1+u}\right)du$$ by integration by parts. The edge terms are zero, so we are left with (integrating the first integrand by parts again) $$\int_0^\infty e^{-u}\ln(1+u)du+\int_0^\infty e^{-u}\frac{u}{1+u}du$$ $$=[-e^{-u}\ln(1+u)]_0^\infty+\int_0^\infty \frac{e^{-u}}{1+u}du+\int_0^\infty e^{-u}\frac{u}{1+u}du$$ $$=0+\int_0^\infty e^{-u}du = 1$$ Thus, combining this with our previous result in the first integral: $$2\int_0^\infty u e^{-u}\ln\left(1+\frac{1}{u}\right)du=2(1-(1-\gamma))=2\gamma$$
Let $ I$ your integral (2) (without the factor $2$). As $1-(x+1)\exp(-x)$ is a primitive of $x\exp(-x)$, an integration by parts gives $$I=\int_0^{+\infty} (\frac {1}{1+x}-\exp(-x))\frac{dx}{x}$$ Then use http://mathworld.wolfram.com/Euler-MascheroniConstant.html
