Integral $\int_0^\infty\frac{1}{x\,\sqrt{2}+\sqrt{2\,x^2+1}}\cdot\frac{\log x}{\sqrt{x^2+1}}\mathrm dx$ I need your assistance with evaluating the integral
$$\int_0^\infty\frac{1}{x\,\sqrt{2}+\sqrt{2\,x^2+1}}\cdot\frac{\log x}{\sqrt{x^2+1}}dx$$
I tried manual integration by parts, but it seemed to only complicate the integrand more. I also tried to evaluate it with a CAS, but it was not able to handle it.
 A: 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
$$
\begin{align*}
\frac{dE'(k)}{dk}&= \frac{k}{k^{'2}}(K'(k)-E'(k))\\
\frac{dK'(k)}{dk}&= \frac{k^2 K'(k)-E^{'}(k)}{kk^{'2}}
\end{align*}
$$
A: $$\frac{1+\ln\sqrt[4]2}{\sqrt{2\,\pi}}\Gamma\left(\frac34\right)^2-\frac{\sqrt{2\,\pi^3}}8\Gamma\left(\frac34\right)^{-2}+(\ln2-1)\,\sqrt2$$
A: This is not a full answer but how far I got, maybe someone can complete
$\frac{1}{x\,\sqrt{2}+\sqrt{2\,x^2+1}}=\frac{1}{\sqrt{2\,x^2}+\sqrt{2\,x^2+1}}=\frac{\sqrt{2\,x^2+1}-\sqrt{2\,x^2}}{\sqrt{2\,x^2+1}^2-\sqrt{2\,x^2}^2}=\sqrt{2\,x^2+1}-\sqrt{2\,x^2}$
So you are looking at
$$\int_0^\infty(\sqrt{\frac{2\,x^2+1}{x^2+1}} - \sqrt{\frac{2\,x^2}{x^2+1}}) \log x\,dx$$
I was then looking at $t = \frac{x^2}{x^2+1}$ but then I don't know
