Let $$I(t) = \int_0^\infty \dfrac{\exp(-x) - \exp(-xt)}{x} \, dx$$ Then
$$\dfrac{dI}{dt} = \int_0^{\infty} \exp(-xt) \, dx = \left. \dfrac{\exp(-xt)}{-t} \right \vert_0^\infty = \dfrac1t$$
Hence, $$I(t) = \ln(t) + c$$ But $$I(1) = \int_0^\infty \dfrac{\exp(-x) - \exp(-x)}{x} \, dx = \int_0^\infty 0 \, dx = 0 \implies c =0$$ Hence, $$I(t) = \ln(t)$$
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
Note
\begin{align} &\int_0^\infty\frac{\ln{x}\ln{(1+ix)}}{1+x^2}dx\\ =&\ \frac12 \int_0^\infty\frac{\ln{x}\ln{(1+x^2)}}{1+x^2}dx + i\int_0^\infty\frac{\ln{x}\tan^{-1}x}{1+x^2}dx \end{align} where \begin{align} \int_0^\infty\frac{\ln{x}\ln{(1+x^2)}}{1+x^2}{dx}& \overset{x\to \frac1x}=\int_0^\infty\frac{\ln^2{x}}{1+x^2}dx =\frac{\pi^3}8\\ \\ \int_0^\infty \frac{\ln x\tan^{-1}x}{1+x^2}dx =& \int_0^\infty \int_0^1 \frac{x\ln x}{(1+x^2)(1+y^2x^2)} \overset{x\to \frac1{xy}}{dx}dy\\ = & \ \frac1{2}\int_0^1\int_0^\infty \frac{-x\ln y}{(1+x^2)(1+{y^2}x^2)} {dx}\ dy\\ =& \ \frac12\int_0^1\frac{\ln^2 y}{1-y^2}dy =\frac78\zeta(3) \end{align}