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.

One possible way is to introduce $$ I(s)=\frac{1}{16}\int_0^{\infty}\frac{y^{s-\frac34}dy}{1+y}.\tag{1}$$ The integral you are looking for is obtained as $I'(0)$ after the change of variables $y=x^4$.

Let us make in (1) another change of variables: $\displaystyle t=\frac{y}{1+y}\Longleftrightarrow y=\frac{t}{1-t},dy=\frac{dt}{(1-t)^2}$. This gives \begin{align} I(s)&=\frac{1}{16}\int_0^1t\cdot\left(\frac{t}{1-t}\right)^{s-\frac74}\cdot \frac{dt}{(1-t)^2}=\\ &=\frac{1}{16}\int_0^1t^{s-\frac34}(1-t)^{-s-\frac{1}{4}}dt=\\& =\frac{1}{16}B\left(s+\frac14,-s+\frac34\right)=\\& =\frac{1}{16}\Gamma\left(s+\frac14\right)\Gamma\left(-s+\frac34\right)=\\ &=\frac{\pi}{16\sin\pi\left(s+\frac14\right)}. \end{align} Differentiating this with respect to $s$, we indeed get $$I'(0)=-\frac{\pi^2\cos\frac{\pi}{4}}{16\sin^2\frac{\pi}{4}}=-\frac{\pi^2\sqrt{2}}{16}.$$

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 $$ \begin{align} \int_{0}^{\infty }\frac {\ln x}{x^4+1}\ dx &=\frac{\pi}{4i}\left(\frac {\ln e^\frac{3i\pi}{4}}{e^\frac{3i\pi}{4}}-\frac {\ln e^\frac{i\pi}{4}}{e^\frac{i\pi}{4}}\right)\\ &=\frac{\pi}{4i}\left(-\frac{i\pi}{2\sqrt{2}}\right)\\ &=\boxed{\color{blue}{-\Large\frac{\pi^2}{16}\sqrt{2}}} \end{align} $$ $$\\$$

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