It is easy to see that the integral is equivalent to
$$
\begin{align*}
\int_0^\infty \frac{1}{x\sqrt{2}+\sqrt{2x^2+1}}\frac{\log x}{\sqrt{1+x^2}}dx &= \sqrt{2}\int_0^\infty \frac{\sqrt{x^2+\frac{1}{2}}-x}{\sqrt{1+x^2}}\log x\; dx\tag{1}
\end{align*}
$$
This integral is a special case of the following generalised equation:
$$\begin{align*}\mathcal{I}(k) :&= \int_0^\infty \frac{\sqrt{x^2+k^2}-x}{\sqrt{1+x^2}}\log x\; dx \\ &= E'(k)-\left(\frac{1+k^2}{2} \right)K'(k)+\left(k^2 K'(k)-E'(k) \right)\frac{\log k}{2}+\log 2-1 \tag{2}\end{align*}$$
where $K'(k)$ and $E'(k)$ are complementary elliptic integrals of the first and second kind respectively.
Putting $k=\frac{1}{\sqrt{2}}$ in equation $(2)$,
$$
\begin{align*}
\mathcal{I}\left(\frac{1}{\sqrt{2}}\right)&=E'\left(\frac{1}{\sqrt{2} }\right)-\frac{3}{4}K'\left(\frac{1}{\sqrt{2}} \right)-\left\{\frac{1}{2} K'\left(\frac{1}{\sqrt{2}} \right)-E'\left(\frac{1}{\sqrt{2}} \right)\right\}\frac{\log 2}{4}+\log 2-1
\end{align*}
$$
Using the special values,
$$
\begin{align*}
E'\left(\frac{1}{\sqrt2} \right) &= \frac{\Gamma\left(\frac{3}{4} \right)^2}{2\sqrt\pi}+\frac{\sqrt{\pi^3}}{4\Gamma\left(\frac{3}{4} \right)^2}\\
K'\left(\frac{1}{\sqrt2} \right) &= \frac{\sqrt{\pi^3}}{2\Gamma\left(\frac{3}{4} \right)^2}
\end{align*}
$$
we get
$$
\mathcal{I}\left(\frac{1}{\sqrt{2}}\right)=\frac{1+\log\sqrt[4]2}{2\sqrt{\,\pi}}\Gamma\left(\frac34\right)^2-\frac{\sqrt{\,\pi^3}}8\Gamma\left(\frac34\right)^{-2}+(\log 2-1)\, \tag{3}
$$
Putting this in equation $(1)$, we get the answer that Cleo posted.
How to prove Equation $(2)$?
We begin with Proposition 7.1 of "The integrals in Gradshteyn and Ryzhik:
Part 16" by Boettner and Moll.
$$\int_0^\infty \frac{\log x}{\sqrt{(1+x^2)(m^2+x^2)}}dx = \frac{1}{2}K'(m)\log m$$
Multiplying both sides by $m$ and integrating from $0$ to $k$:
$$
\begin{align*}
\int_0^\infty \frac{\sqrt{x^2+k^2}-x}{\sqrt{1+x^2}}\log x\; dx &= \frac{1}{2}\int_0^k m K'(m)\log(m)\; dm
\end{align*}
$$
The result follows since
$$\begin{align*} \int m K'(m)\log(m)\; dm &= 2E'(m)-\left(1+m^2 \right)K'(m)+\left(m^2 K'(m)-E'(m) \right)\log m\\ &\quad +\text{constant} \tag{4}
\end{align*}$$
One can verify equation $(4)$ easily by differentiating both sides with respect to $m$ and using the identities
$$
\begin{align*}
\frac{dE'(k)}{dk}&= \frac{k}{k^{'2}}(K'(k)-E'(k))\\
\frac{dK'(k)}{dk}&= \frac{k^2 K'(k)-E^{'}(k)}{kk^{'2}}
\end{align*}
$$
Best Answer
A little late to the party here, but I finally managed to get a solution.
From the answer here, and using a trivial rescaling of the integration variable, we can see that:
$$f(a,t)=\int_{0}^{\infty}\frac{dx}{x^2+a}e^{-\frac{x^2}{2t}}=\frac{\pi}{2\sqrt{a}}\text{erfc}\Big(\sqrt{\frac{a}{2t}}\Big)\exp\Big(\frac{a}{2t}\Big)$$
As a function of $a$, the above has an integrable singularity at $a=0$ and is continuous, and therefore the quantity $\int_{0}^{b}f(a,t)da$ converges for any $b \geq 0$ in the generalized integral sense. Also, we observe that the integral $$I(a,t)=\int_{0}^{\infty}dx\ln(x^2+a)e^{-\frac{x^2}{2t}}$$ converges for all $a\geq 0$ as well. Hence we can safely write that:
$$I(a,t)-I(0,t)=\int_{0}^{\infty}dx ~e^{-x^2/2t}\int_{0}^{a}\frac{da'}{x^2+a'}\Rightarrow \\I(a,t)=2\int_{0}^{\infty}dx\ln x ~e^{-x^2/2t}+\pi\int_{0}^{a}\frac{d\alpha}{2\sqrt{\alpha}}\text{erfc}\Big(\sqrt{\frac{\alpha}{2t}}\Big)\exp\Big(\frac{\alpha}{2t}\Big)=I_1+I_2$$
For the first integral we set $u=\frac{x^2}{2t}$ and we obtain
$$I_1(t)=\frac{\sqrt{2t}}{2}\Big[\int_0^{\infty}du \ln u~u^{-1/2}e^{-u}-\ln (1/2t)\int_0^{\infty}du\frac{e^{-u}}{\sqrt{u}}\Big]=\frac{1}{2}\sqrt{2\pi t}(\psi^{(0)}(1/2)+\ln(2t))=\sqrt{\frac{\pi t}{2}}(\ln(2t)-2\ln2-\gamma)$$
For the second we set $u=\sqrt{\frac{\alpha}{2t}}$ instead and we obtain that:
$$I_2=\pi \sqrt{2t}\int_{0}^{\sqrt{\frac{a}{2t}}}du ~e^{u^2}\text{erfc}(u)$$
and using $\text{erfc}(x)=1-\text{erf}(x)$ and the fact that $(\text{erfi})'(x)=\frac{2}{\sqrt{\pi}}e^{x^2}$ we obtain the following handsome looking final expression, after a few simplifications:
$$I=\sqrt{\frac{\pi t}{2}}\Big[\ln(t/2)-\gamma+\pi~\text{erfi}(\sqrt{\frac{a}{2t}})-2\sqrt{\pi}\int_{0}^{\sqrt{\frac{a}{2t}}}e^{u^2}\text{erf}(u)du\Big]$$
This is where things actually get interesting. Consider the function $f(x)=e^{x^2}\int_{0}^{x}e^{-u^2}du$, and perform $M$ integration by parts on it:
$$f(x)=\sum_{n=0}^M \frac{2^n x^{2n+1}}{1\cdot 3\cdot 5...(2n+1)}-e^{x^2}\frac{2^{M+1}}{1\cdot 3\cdot 5...(2M+1)}\int_{0}^x{t^{2M+2}e^{-t^2}dt}$$
Taking $M\to\infty$, which we can do for any $x$, by a simple estimate of the remainder above, we recover the Taylor series:
$$f(x)=\sum_{n=0}^{\infty}\frac{2^n x^{2n+1}}{1\cdot 3\cdot 5...(2n+1)}$$
Now by integrating and using the definition of hypergeometric functions we find:
$$\int_{0}^t f(x)dx=\sum_{n=0}^{\infty}\frac{2^n t^{2n+2}}{1\cdot 3\cdot 5...(2n+1)(2n+2)}=t^2 ~_2F_2(1,1;\frac{3}{2},2|t^2)$$
for the coveted result:
$$I(a,t)=\sqrt{\frac{\pi t}{2}}\Big[\ln(t/2)-\gamma+\pi~\text{erfi}(\sqrt{\frac{a}{2t}})-\frac{a}{t} ~_2F_2(1,1;\frac{3}{2},2|\frac{a}{2t})\Big]$$
which confirms the Mathematica evaluation of $I(t^2,t)$, which is the desired integral, from the comments above.