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
$$
\begin{align*}
\frac{dE'(k)}{dk}&= \frac{k}{k^{'2}}(K'(k)-E'(k))\\
\frac{dK'(k)}{dk}&= \frac{k^2 K'(k)-E^{'}(k)}{kk^{'2}}
\end{align*}
$$
We have the following closed form.
Proposition. $$
\int_0^1\log \left(1-x\right)\log \left(-\log x\right)\:dx=\gamma-\gamma_1+\gamma_1(1,0)\tag1
$$
where $\displaystyle \gamma$ is the Euler-Mascheroni constant, where $\gamma_1$ is the Stieltjes constant,
$$\gamma_1 = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log n}n-\int_1^N\frac{\log t}t\:dt\right)$$
and where $\gamma_1(a,b)$ is the poly-Stieltjes constant (see here),
$$\gamma_1(a,b) = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log (n+a)}{n+b}-\int_1^N\frac{\log t}t\:dt\right)\!.$$
Proof.
One may recall the classic integral representation of the Euler gamma function
$$
\frac{\Gamma(s)}{(a+1)^s}=\int_0^\infty t^{s-1} e^{-(a+1)t}\:dt, \qquad s>0,\, a>-1. \tag2
$$ By differentiating $(2)$ with respect to $s$, putting $s=1$ and making the change of variable $x=e^{-t}$, we get
$$
\int_0^1x^a\log\left(-\log x\right)\:dx=-\frac{\gamma+\log(a+1)}{a+1},\qquad a>-1, \tag3
$$
where $\displaystyle \gamma$ is the Euler-Mascheroni constant. We are allowed to insert the standard Taylor series expansion,
$$
\log (1-x)= -\sum_{n=1}^{\infty} \frac{x^n}n, \qquad |x|<1,\tag4
$$ into the given integral, then using $(3)$ we obtain
$$
\begin{align}
\int_0^1\log \left(1-x\right)\log \left(-\log x\right)\:dx&=-\int_0^1\sum_{n=1}^{\infty}\frac{x^n}n \:\log (-\log x)\:dx\\
&=-\sum_{n=1}^{\infty} \frac1n\int_0^1 x^n\log (-\log x)\:dx\\
&=\sum_{n=1}^{\infty} \frac{\gamma+\log(n+1)}{n(n+1)}\\
&=\gamma \sum_{n=1}^{\infty}\frac1{n(n+1)}+\sum_{n=1}^{\infty} \frac{\log(n+1)}{n(n+1)}\\
&=\gamma +\sum_{n=1}^{\infty} \frac{\log(n+1)}{n(n+1)},\tag5
\end{align}
$$ and we may conclude with Theorem $2$ here to get
$$
\begin{align}
\sum_{n=1}^{\infty} \frac{\log (n+1)}{n(n+1)}=\gamma_1({1,0})-\gamma_1,\tag6
\end{align}
$$
since $\gamma_1(1,1)=\gamma_1$.
Remark. I am inclined to believe that the poly-Stieltjes constants will turn out to be a tool for many of the considered integrals.
Best Answer
Let's apply the substitution $\sqrt{x^4+1}-x^2=\sqrt{t}$: $$x^2=\frac{1-t}{2\sqrt{t}},\quad dx=-\frac{\sqrt2}{8}t^{-5/4}(t+1)(1-t)^{-1/2}dt,$$ $$\int_0^\infty\left(\sqrt{x^4+1}-x^2\right)\,dx=\frac{\sqrt2}{8}\left(\int_0^1 t^{1/4}(1-t)^{-1/2}\,dt+\int_0^1 t^{-3/4}(1-t)^{-1/2}\,dt\right)=$$ $$=\frac{\sqrt2}{8}\left(\mathrm{B}\left(\frac54,\frac12\right)+\mathrm{B}\left(\frac14,\frac12\right)\right)=\frac{\Gamma^2(\frac14)}{6\sqrt{\pi}}.$$