When I do $$e^u=x+\sqrt{x^2+1}$$ in (17) page 9, see here, after some computations then I obtain $$\frac{\zeta(3)}{10}=\int_{0}^{\log(\phi)}\frac{u^2(e^{2u}+1)}{e^{2u}-1}du,$$ where $\phi=\frac{1+\sqrt{5}}{2}$ is the golden ratio and $\zeta(3)$ is the Apéry's constant.
I don't know if this exercise is in the literature:
Question. Can you prove what I say? This is, can you show $$\frac{\zeta(3)}{10}=\int_{0}^{\log(\phi)}\frac{u^2(e^{2u}+1)}{e^{2u}-1}du?$$ Then I can check if my computations were rights. Thanks in advance.