calculus - 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.

Answer

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
$$I'(0)=-\frac{\pi^2\cos\frac{\pi}{4}}{16\sin^2\frac{\pi}{4}}=-\frac{\pi^2\sqrt{2}}{16}.$$