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.

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
