By reading a german wikipedia (see here) about integrals, i stumpled upon this entry
(Click for the source) Where the result was proven using complex analysis. Is there any method to show the equality using real methods? Any help will be appreciated =)
By setting $e^\eta=v=1-u$ and exploiting the inverse Laplace transform we have:
$$\int_{1}^{+\infty}\frac{dx}{(x+v)\left(\pi^2+\log^2 x\right)}=\frac{1}{\pi}\int_{1}^{+\infty}\frac{dx}{x+v}\int_{0}^{+\infty}\sin(\pi z)\,x^{-z}\,dz.\tag{1}$$ Moreover, if $0<z<1$, by exploiting the Euler beta function and the reflection formulas for the $\Gamma$ function we have: $$ \int_{0}^{+\infty}\frac{x^{-z}}{x+v}\,dx = \frac{\pi}{v^z\sin(\pi z)}=\int_{0}^{+\infty}\frac{x^z}{x+vx^2}\,dx\tag{2}$$ and rearranging carefully we get:
$$ \int_{0}^{+\infty}\frac{dx}{(x+v)\left(\pi^2+\log^2 x\right)}=\int_{0}^{1}t^{-v}\,dt - \int_{0}^{+\infty}v^{-z}\,dz = \frac{1}{1-v}-\frac{1}{\log v}\tag{3}$$ as wanted. Anyhow, this is not the sketch of a really alternative proof, since the inverse Laplace transform is just the residue theorem in disguise.
I'm not sure about the full solution, but there is a way to find an interesting functional equation for this integral.
First, let's get rid of the silly restriction on $u$. By numerical evaluation, the integral exists for all $u \in (-\infty,1)$
Now let's introduce the more convenient parameter:
$$v=1-u$$
$$I(v)=\int_0^{\infty} \frac{dx}{(v+x)(\pi^2+\ln^2 x)}$$
Now let's make a change of variable:
$$x=e^t$$
$$I(v)=\int_{-\infty}^{\infty} \frac{e^t dt}{(v+e^t)(\pi^2+t^2)}$$
$$I(v)=\int_{-\infty}^{\infty} \frac{(v+e^t) dt}{(v+e^t)(\pi^2+t^2)}-v \int_{-\infty}^{\infty} \frac{ dt}{(v+e^t)(\pi^2+t^2)}=1-v J(v)$$
Now let's make another change of variable:
$$t=-z$$
$$I(v)=\int_{-\infty}^{\infty} \frac{e^{-z} d(-z)}{(v+e^{-z})(\pi^2+z^2)}=\int_{-\infty}^{\infty} \frac{ dz}{(1+v e^z)(\pi^2+z^2)}=\frac{1}{v} J \left( \frac{1}{v} \right)$$
Now we get:
$$1-v J(v)=\frac{1}{v} J \left( \frac{1}{v} \right)=I(v)$$
$$v J(v)+\frac{1}{v} J \left( \frac{1}{v} \right)=1$$
$$v \in (0,\infty)$$
For example, we immediately get the correct value:
$$J(1)=I(1)=\int_0^{\infty} \frac{dx}{(1+x)(\pi^2+\ln^2 x)}=\frac{1}{2}$$
We can also check that this equation works for the known solution (which is actually valid on the whole interval $v \in (0,\infty)$, except for $v=1$).
$$I(v)=\frac{1}{1-v}+\frac{1}{\ln v}$$
$$J(v)=-\frac{1}{1-v}-\frac{1}{v \ln v}$$
$$J \left( \frac{1}{v} \right)=\frac{v}{1-v}+\frac{v}{\ln v}$$
$$1-v J(v)=\frac{1}{v} J \left( \frac{1}{v} \right)$$
Now this is not a solution of course (except for $I(1)$), but it's a big step made without any complicated integration techniques.
Basically, if we define:
$$f(v)=vJ(v)$$
$$I(v)=1-f(v)$$
We need to solve a simple functional equation:
$$I(v)+I \left( \frac{1}{v} \right)=1$$
Below is a real method by utilizing $\int_0^1 \sin(\pi t)x^{-t}dt=\frac{\pi(x+1)}{x(\ln^2 x+\pi^2)}$
\begin{align} I=&\int_0^\infty \frac{1}{(x+1-u)(\ln^2 x+\pi^2)} \ dx\\ =& \int_0^\infty \frac{1}{x+1-u}\bigg( \frac1\pi\int_0^1 \sin(\pi t)x^{-t}dt -\frac1{x(\ln^2 x+\pi^2)}\bigg)dx\\ =&\ \frac1\pi\int_0^1\sin(\pi t) \left(\int_0^\infty\frac{x^{-t}}{x+1-u}dx\right) dt\\ &\hspace{10mm}-\frac1{1-u}\int_0^\infty\left(\frac1x-\frac1{x+1-u}\right)\frac1{\ln^2 x+\pi^2}dx\\ =& \int_0^1 (1-u)^{-t}dt-\frac1{1-u}+\frac1{1-u}I\\ =& \ \frac{1-u}{-u}\left( \frac {-u}{(1-u)\ln(1-u)}-\frac1{1-u}\right)\\ =&\ \frac{1}{u}+\frac{1}{\ln(1-u)} \end{align} where $\int_0^\infty\frac{x^{-t}}{x+1-u}dx=\frac{\pi(1-u)^{-t}}{\sin(\pi t)}$ and $ \int_0^\infty \frac1{x(\ln^2 x+\pi^2)}dx= 1$ are used in above evaluation.
