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
The integral is somehow nontrivial. I would proceed as follows.
OP's second attempt seems to be on the right track (at least for my approach). Denote the integral as $I$, substitute $\dfrac{x^4-1}{2x^2}=t$ so that $$ I=\int_0^\infty\frac{dx}{\sqrt{1+\exp\left(\dfrac\pi2\left(x^2-\dfrac1{x^2}\right)\right) }}=\frac12\int_{-\infty}^\infty\frac{\sqrt{\sqrt{t^2+1}+t}}{ \sqrt{t^2+1}}\frac{dt}{\sqrt{e^{\pi t}+1}} $$ The term $\dfrac{\sqrt{\sqrt{t^2+1}+t}}{ \sqrt{t^2+1}}$ reminds me of $\sin(\theta/2)$, so I came up with $$ \int_{-\infty }^{\infty } \frac{1}{\left(v^2-t\right)^2+1} dv=\Im \int_{-\infty }^{\infty } \frac{1}{v^2-t-i} dv =\Im\frac{\pi}{\sqrt{-t-i}} =\frac\pi{\sqrt2}\dfrac{\sqrt{\sqrt{t^2+1}+t}}{ \sqrt{t^2+1}} $$ and the integral becomes $$ I=\frac1{\sqrt2\pi}\int_{-\infty }^{\infty }\int_{-\infty}^\infty \frac{1}{\left(v^2-t\right)^2+1}\frac{1}{\sqrt{e^{\pi t}+1}}dvdt\\ =\frac1{\sqrt2\pi}\int_{-\infty }^{\infty }du\int_{-\infty}^\infty \frac{1}{\left(u^2(e^{\pi t}+1)-t\right)^2+1}dt \tag{$\ast$} $$ where $v=u\sqrt{e^{\pi t}+1}$.
Now consider the meromorphic function $f(z)=\dfrac1{u^2(e^{\pi z}-1)+z}$ and integrate on the rectangular contour $[-R,R]\times [-1,1]$. When $R\to\infty$, it is easy to see that the integrals on the vertical lines tend to vanish as $e^{-\pi R},\Re(z)=R\to\infty$ and $R^{-1}, \Re(z)=-R\to -\infty$. Meanwhile, the two on the horizontal line combines to get $$ \oint f(z)dz=\int_{-\infty}^\infty f(t-i)dt+\int^{-\infty}_\infty f(t+i)dt=2i\int_{-\infty}^\infty \frac{1}{\left(u^2(e^{\pi t}+1)-t\right)^2+1}dt $$ Thus, it suffice to investigate the location of the poles of $f(z)$. Though it normally has infinite poles on the complex plane, I claim that there's only one in the contour, namely $z=0$. To see this, let $z=\Re(z)+i \Im(z)=\xi+i\eta$ be a pole, then take the real and imaginary part $$ \cases{ u^2(e^{\pi\xi}\cos(\pi\eta)-1)+\xi=0 &(r)\\[5pt] u^2e^{\pi \xi}\sin(\pi \eta)+\eta=0 &(i) } $$ The two terms on the LHS of equation (i) always has the same sign when $-\pi<\eta<\pi$, so its only solution in the contour is $\eta=0$. Now equation (r) becomes $u^2(e^{\pi\xi}-1)+\xi=0$. It is easy to see that LHS is a monotonic function of $\xi$ and has a root $\xi=0$, so it is the only one, proving the claim.
By the residue theorem, $$ \oint f(z)dz=2\pi i\text{Res}[f(z),0]=\frac{2\pi i}{1+\pi u^2} $$ that is $$ \int_{-\infty}^\infty \frac{1}{\left(u^2(e^{\pi t}+1)-t\right)^2+1}dt=\frac{\pi}{1+\pi u^2} $$ plug this into $(\ast)$ $$ I=\frac1{\sqrt2\pi}\int_{-\infty}^\infty \frac{\pi}{1+\pi u^2}du=\sqrt{\frac{\pi}2} $$ as desired.
Update
I played around with my method and found more interesting result as one can plug any analytic function into $f(z)$ that keeps the contour integral neet. For example, $$ \int_0^\infty \frac{dx}{\sqrt{\cosh(x\sqrt{x^2+\pi})}}=\frac{\pi}{2\sqrt2} $$ Denote $t(x)=\dfrac{x^4-1}{2x^2}$ and $$ J_n=\int_0^\infty\frac{dx}{\sqrt{1+e^{\pi t(x)}}^{2n+1}}=\sqrt{\frac{\pi }{2}}\cdot[z^{-1}]\frac{1}{\sqrt{z \left(e^{ z}-1\right)} \left(1-e^{ z}\right)^n} $$ is always an rational multiple of $\sqrt{2\pi}$, like $$ J_1=\frac{3 }{8}\sqrt{2\pi}\quad J_2=\frac{65}{192}\sqrt{2\pi}\quad J_3=\frac{245 }{768}\sqrt{2\pi}\quad \cdots $$ One with more poles in the contour $$ \int_0^\infty\frac{\operatorname{sech}(\pi t(x))}{\sqrt{1+e^{\pi t(x)}}}dx=\sqrt{\sqrt{2}+1}-\sqrt{\frac{\pi }{2}} $$
However, now some of them seem to be trivial under @Sangchul Lee's brilliant approach.