Using the Eulerian integrals evaluate $\int_0^\infty \frac{\ln^2{x}}{1+x^4} \mathrm{d}x$ The question asks to evaluate the integral: $$\int_0^\infty \frac{\ln^2{x}}{1+x^4} \mathrm{d}x.$$
I have tried a few substitutions but am not getting anywhere.
Thanks in advance!
 A: Assume $0<s<4$. We start with the following general result

$$
\int_0^\infty \frac{x^{s-1}}{1+x^4}\mathop{dx}=\frac{\pi}4\:\frac1{\sin\left(\pi s/4\right)}. \tag1
$$

Proof.
By performing the change of variable $u=\dfrac{1}{1+x^4}$, then using the Euler beta function $B$, we have
$$
\begin{align}
\int_0^\infty \frac{x^{s-1}}{1+x^4}\mathop{dx}&=\frac14\int_0^1 u^{\large 1-\frac{s}4-1}(1-u)^{\large \frac{s}4-1}\mathop{du}\\\\
&=\frac14B\left(1-\frac s4,\frac s4\right)\\\\
&=\frac14\Gamma\left( 1-\frac{s}4\right)\Gamma\left(\frac{s}4\right)\\\\
&=\frac{\pi}4\:\frac1{\sin\left(\pi s/4\right)}
\end{align}
$$ giving $(1)$, where we have used the reflection formula for the $\Gamma$ function.
Then one may differentiate $(1)$ twice getting
$$
\int_0^\infty \frac{x^{s-1}(\ln x)^2}{1+x^4}\mathop{dx}=\frac{\pi}4\left(\frac1{\sin\left(\pi s/4\right)}\right)_{\large s}'' \tag2
$$ then put $s:=1$ to obtain

$$
\int_0^\infty \frac{\ln^2 x}{1+x^4}\mathop{dx}=\frac{3\pi^3}{64}\sqrt{2}. \tag3
$$

A: Note
\begin{eqnarray}
\int_0^\infty \frac{\ln^2{x}}{1+x^4} \mathrm{d}x&=&\int_0^1 \frac{\ln^2{x}}{1+x^4} \mathrm{d}x+\int_1^\infty \frac{\ln^2{x}}{1+x^4} \mathrm{d}x\\
&=&\int_0^1 \frac{(1+x^2)\ln^2{x}}{1+x^4} \mathrm{d}x\\
&=&\frac{3\pi^3}{64}\sqrt2.
\end{eqnarray}
See the result in this post.
