Evaluating $\int_0^\infty \frac{1}{x+1-u}\cdot \frac{\mathrm{d}x}{\log^2 x+\pi^2}$ using real methods. By reading a german wikipedia (see here) about integrals, i stumpled upon this entry

27 1.5
  $$ \color{black}{
\int_0^\infty \frac{1}{x+1-u}\cdot \frac{\mathrm{d}x}{\log^2 x+\pi^2} =\frac{1}{u}+\frac{1}{\log(1-u)}\,, \qquad u \in (0,1)}
$$
  

(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 =)
 A: 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. 
A: 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:

$$I(v)+I \left( \frac{1}{v} \right)=1$$

