Show that $\int_{0}^{\infty }\frac {\ln x}{x^4+1}\ dx =-\frac{\pi^2 \sqrt{2}}{16}$ I could prove it using the residues but I'm interested to have it in a different way (for example using Gamma/Beta or any other functions) to show that
$$
\int_{0}^{\infty}\frac{\ln\left(x\right)}{x^{4} + 1}\,{\rm d}x
=-\frac{\,\pi^{2}\,\sqrt{\,2\,}\,}{16}.
$$
Thanks in advance.

Another approach, we can split the denominator part as follows
$$
\frac{1}{x^4+1}=\frac{1}{2i}\left(\frac{1}{x^2-i}-\frac{1}{x^2+i}\right).
$$
Consequently, the integral becomes
$$
\int_{0}^{\infty }\frac {\ln x}{x^4+1}\ dx =\frac{1}{2i}\int_{0}^{\infty }\left(\frac{\ln x}{x^2-i}-\frac{\ln x}{x^2+i}\right)\ dx.
$$
Using formula from here,
$$
\int_0^{\infty}\frac{\ln x}{x^2+a^2}\ dx=\frac {\pi \ln a}{2a},
$$
we obtain
$$
\begin{align}
\int_{0}^{\infty }\frac {\ln x}{x^4+1}\ dx&=\frac{1}{2i}\left(\frac {\pi \ln \sqrt{-i}}{2\sqrt{-i}}-\frac {\pi \ln \sqrt{i}}{2\sqrt{i}}\right)\\
&=\frac{\pi}{4i}\left(\frac {\ln i^{\frac{3}{2}}}{i^{\frac{3}{2}}}-\frac {\ln i^{\frac{1}{2}}}{i^{\frac{1}{2}}}\right).
\end{align}
$$
Taking $0\le\theta\le2\pi$, from Euler's formula we have
$$
e^\frac{i\pi}{2}=\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}=i.
$$
Thus
$$
\begin{align}
\int_{0}^{\infty }\frac {\ln x}{x^4+1}\ dx
&=\frac{\pi}{4i}\left(\frac {\ln e^\frac{3i\pi}{4}}{e^\frac{3i\pi}{4}}-\frac {\ln e^\frac{i\pi}{4}}{e^\frac{i\pi}{4}}\right)\\
&=\frac{\pi}{4i}\left(-\frac{i\pi}{2\sqrt{2}}\right)\\
&=\boxed{\color{blue}{-\Large\frac{\pi^2}{16}\sqrt{2}}}
\end{align}
$$
$$\\$$

$$\large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$
