Assume that $a>0$.
Very similar to my answer here, let $$I(a) = \int_{0}^{\infty} \frac{\cos(ax)-e^{-{ax}}}{x(1+x^{4})} \, \ln(x) \, \mathrm dx.$$
Then $$ \begin{align} I^{(4)}(a) + I(a) &= \int_{0}^{\infty} \frac{\cos(ax)-e^{-ax}}{x} \, \ln(x) \, \mathrm dx \\ &= \int_{0}^{\infty} \frac{\cos(u) -e^{-u}}{u} \, \ln(u) \, \mathrm du -
\int_{0}^{\infty} \frac{\cos(u) -e^{-u}}{u} \, \ln(a) \, \mathrm du \\ &\overset{(1)}= \int_{0}^{\infty} \frac{\cos(u) -e^{-u}}{u} \, \ln(u) \, \mathrm du -0\\ &=\lim_{s \to 0^{+}} \frac{\mathrm d}{\mathrm ds}\int_{0}^{\infty} \left(\cos(u) - e^{-u} \right) u^{s-1} \, \mathrm du \\ &= \lim_{s \to 0^{+}}\frac{\mathrm d}{\mathrm ds} \, \Gamma(s) \left( \cos \left(\frac{\pi s}{2} \right)-1\right) \\ &= \lim_{s \to 0^{+}} \left(\Gamma'(s) \left(\cos \left(\frac{\pi s}{2}\right)-1 \right)- \Gamma(s)\frac{\pi}{2} \sin \left(\frac{\pi s}{2} \right) \, \right) \\ &= \lim_{s \to 0^{+}} \left(\left(- \frac{1}{s^{2}} + O(1) \right)\left(\cos \left(\frac{\pi s}{2}\right)-1 \right) - \frac{\pi}{2}\left(\frac{1}{s} +O(1) \right)\sin \left(\frac{\pi s}{2} \right)\right) \\ &= \frac{\pi^{2}}{8} - \frac{\pi^{2}}{4} \\ &= - \frac{\pi^{2}}{8}. \end{align}$$
The general solution of the above linear differential equation with constant coefficients is $$ \small I(a) = C_{1}e^{a / \sqrt{2}} \cos \left(\frac{a}{\sqrt{2}} \right) +C_{2}e^{a / \sqrt{2}} \sin \left(\frac{a}{\sqrt{2}} \right) +C_{3} e^{-a / \sqrt{2}} \cos \left(\frac{a}{\sqrt{2}} \right) +C_{4} e^{-a / \sqrt{2}} \sin \left(\frac{a}{\sqrt{2}} \right) - \frac{\pi^{2}}{8}.$$
The initial conditions are $\lim_{a \to 0^{+}} I(a) = 0$ and $$\lim_{a \to 0^{+}} I'(a) = \int_{0}^{\infty} \frac{\ln(x)}{1+x^{4}} \, \mathrm dx = \lim_{s \to 1} \frac{\mathrm d}{\mathrm ds} \int_{0}^{\infty} \frac{x^{s-1}}{1+x^{4}} \, \mathrm dx =\lim_{s \to 1} \frac{\mathrm d}{\mathrm ds} \frac{\pi}{4} \, \csc \left(\frac{\pi s}{4} \right) = -\frac{\pi^{2}}{8 \sqrt{2}}.$$
Since $I(a)$ remains finite as $a \to + \infty$, $C_{1}$ and $C_{2}$ must be zero.
And the initial condition $\lim_{a \to 0^{+}} I(a) =0$ means that $C_{3}= \frac{\pi^{2}}{8}$.
Finally, to satisfy the initial condition $\lim_{a \to 0^{+}} I'(a) = -\frac{\pi^{2}}{8 \sqrt{2}}$, $C_{4}$ must be zero.
Therefore, $$I(a) = \frac{\pi^{2}}{8} e^{-a / \sqrt{2}} \cos \left(\frac{a}{\sqrt{2}} \right) - \frac{\pi^{2}}{8},$$
and $$ \int_{0}^{\infty} \frac{e^{-x}-\cos(x)}{x (1+x^{4})} \, \ln(x) \, \mathrm dx = -I(1) = \frac{\pi^{2}}{8} \left(1- e^{-1 / \sqrt{2}} \cos \left(\frac{1}{\sqrt{2}} \right)\right). $$
$(1)$ See here for a way to show that the integral vanishes without using complex analysis or special functions.
Evaluate $I_2$ and $I_3$ together, i.e.
\begin{align}
&\int_0^\infty\frac{(6-x^2)\cos x}{4+x^4}dx\\
=& \ \frac12\int_{-\infty}^\infty\frac{(6-x^2)\cos x}{4+x^4}dx\\
=& \ \frac12\int_{-\infty}^\infty\cos x\bigg( \frac{{2x+3}}{\underset{t=x+1}{(x+1)^2+1}}- \frac{{2x+3}}{\underset{t=x-1}{(x-1)^2+1}}\bigg) dx\\
=&\int_{-\infty}^\infty\frac{2t\sin t \sin1+\cos t\cos1}{t^2+1}dt
\end{align}
where the odd terms resulting from the substitutions $t=x\pm 1$ are discarded because they vanish upon integration. Then, utilize the known integrals
$$\int_{-\infty}^\infty \frac{t\sin t}{t^2+1} dt = \int_{-\infty}^\infty \frac{\cos t}{t^2+1} dt= \frac\pi e$$ to arrived at the desired result
$$\int_{-\infty}^{\infty} \frac{\cos x}{\left(1 + x + x^2\right)^2 + 1} \,dx = \frac{\pi }{5e}(1 + 2\sin1 + \cos1)
$$
Best Answer
We can make use of the following result:$$\boxed{\operatorname{Ci}^2(x)+\operatorname{si}^2(x)=\int_0^\infty \frac{e^{-xy}\ln(1+y^2)}{y}dy}$$
$$I_1=\int_0^\infty x\left(\operatorname{Ci}^2(x)+\operatorname{si}^2(x)\right)\operatorname{Ci}(x)dx$$ $$=\int_0^\infty \int_0^\infty \frac{xe^{-xy}\ln(1+y^2)\operatorname{Ci}(x)}{y}dydx$$ $$=\int_0^\infty \frac{\ln(1+y^2)}{y}\left(\int_0^\infty x e^{-xy}\operatorname{Ci}(x)dx\right)dy$$ $$=\int_0^\infty \frac{\ln(1+y^2)}{y}\left(-\frac{d}{dy}\int_0^\infty e^{-xy}\operatorname{Ci}(x)dx\right)dy$$ $$=\frac12\int_0^\infty \frac{\ln(1+y^2)}{y}\frac{d}{dy}\left(\frac{\ln(1+y^2)}{y}\right)dy=\frac14\left(\frac{\ln(1+y^2)}{y}\right)^2\bigg|_0^\infty =0$$ Above the Laplace transform of the cosine integral was used.
Similarly, for $I_2$ we have: $$I_2=\int_0^\infty \frac{\ln(1+y^2)}{y}\left(\int_0^\infty x e^{-xy}\operatorname{si}(x)dx\right)dy$$ $$=\int_0^\infty \frac{\ln(1+y^2)}{y}\left(-\frac{d}{dy}\int_0^\infty e^{-xy}\operatorname{si}(x)dx\right)dy$$ $$=\int_0^\infty \frac{\ln(1+y^2)}{y}\frac{d}{dy}\left(\frac{\arctan y}{y}\right)dy$$ $$=\int_0^\infty \left(\frac{\ln(1+y^2)}{y^2(1+y^2)}-\color{blue}{\frac{\arctan y\ln(1+y^2)}{y^3}}\right)dy$$ $$\overset{\color{blue}{IBP}}=\int_0^\infty \left(\frac12\frac{\ln(1+y^2)}{y^2}-\frac12\frac{\ln(1+y^2)}{1+y^2}-\color{red}{\frac{\arctan y}{y(1+y^2)}}\right)dy$$ $$\overset{\color{red}{IBP}}=\int_0^\infty \left(\frac12\underbrace{\frac{\ln(1+y^2)}{y^2}}_{\pi}-\underbrace{\frac{\ln(1+y^2)}{1+y^2}}_{\pi\ln 2}+\underbrace{\frac{\ln y}{1+y^2}}_{0}\right)dy=\frac{\pi}{2}-\pi\ln 2$$ This time the Laplace transform of the sine integral was used.
Proof for the mentioned result
Since $2\cos x= e^{ix}+e^{-ix}$, we can write: $$\operatorname{Ci}(x)=-\int_x^\infty \frac{\cos t}{t}dt=-\frac12\int_x^\infty \frac{e^{it}}{t}dt-\frac12\int_x^\infty \frac{e^{-it}}{t}dt$$ $$=-\frac12\int_x^\infty \int_0^\infty e^{it}e^{-ty}dydt-\frac12\int_x^\infty \int_0^\infty e^{-it}e^{-ty}dydt$$ $$=-\frac12 \int_0^\infty\int_x^\infty e^{-(y-i)t}dtdy-\frac12\int_0^\infty\int_x^\infty e^{-(y+i)t}dtdy$$ $$=-\frac12 e^{ix} \int_0^\infty \frac{e^{-xy}}{y-i}dy-\frac12 e^{-ix} \int_0^\infty \frac{e^{-xy}}{y+i}dy$$ By the same approach, using $2i\sin x= e^{ix}-e^{-ix}$, we arrive at: $$\operatorname{si}(x)=-\frac1{2i} e^{ix} \int_0^\infty \frac{e^{-xy}}{y-i}dy+\frac1{2i} e^{-ix} \int_0^\infty \frac{e^{-xy}}{y+i}dy$$
Finally, writting $2(a^2+b^2)$ as $(a+b)^2+(a-b)^2$ yields: $$\small \operatorname{Ci}^2(x)+\operatorname{si}^2(x)=\int_0^\infty \frac{e^{-xy}}{y-i}dy\int_0^\infty \frac{e^{-xz}}{z+i}dz\overset{y+z\to y}=\int_0^\infty \int_z^\infty \frac{e^{-xy}}{(y-z-i)(z+i)}dydz$$ $$\small =\int_0^\infty e^{-xy} \int_0^y \frac{1}{(y-z-i)(z+i)}dzdy=\int_0^\infty \frac{e^{-xy}\ln(1+y^2)}{y}dy$$