Show that $\int_{0}^\infty \frac{1}{(1+x-u)(\pi^2+\ln^2(x))}dx=\frac{1}{u}+\frac{1}{\ln(1-u)}$ I am trying to show that $$I=\int_{0}^\infty \frac{1}{(1+x-u)(\pi^2+\ln^2(x))}dx=\frac{1}{u}+\frac{1}{\ln(1-u)}$$
My first thought was to follow the $m=\frac{1}{x}$ which led to
$$I=1+(u-1)\int_{-\infty}^\infty \frac{1}{(e^{-m}+(1-u))(\pi^2+m^2)}dm $$
But that unfortunately did not seem to go anywhere. I original saw this integral in the Laplace Transform section of Jack D'Aurizio's notes, but I, for the life of me, could not seem to use the Laplace Transform to evaluate the integral. I'm also aware that this is very similar to the integral involving the Gregory Coefficients, but I want to avoid those if I can. I've attempted to use differentiating under the integral but couldn't find any reasonable way of doing it. I've tried series, but that did not work either. Any push in the right direction would be much appreciated.
 A: Not an answer, just some progress on the integral
We define
$$P(u)=\int_0^\infty \frac{dx}{(x+u)(\pi^2+\ln^2x)}$$
which is
$$P(u)=1-u\int_{-\infty}^{\infty}\frac{dx}{(e^x+u)(\pi^2+x^2)}.$$
Then preform the sub $x\mapsto \pi\tan x$ so that 
$$\begin{align}
P(u)&=1-\frac{u}\pi\int_{-\pi/2}^{\pi/2}\frac{dx}{u+e^{\pi\tan x}}\\
&=1-\frac{u}\pi\left[\int_{0}^{\pi/2}\frac{dx}{u+e^{\pi\tan x}}+\int_{-\pi/2}^0\frac{dx}{u+e^{\pi\tan x}}\right]\\
&=1-\frac{u}\pi\left[\int_{0}^{\pi/2}\frac{dx}{u+e^{\pi\tan x}}+\int_0^{\pi/2}\frac{dx}{u+e^{-\pi\tan x}}\right]\\
&=1-\frac{u}\pi\int_{0}^{\pi/2}\frac{2u+2\cosh(\pi\tan x)}{1+u^2+2u\cosh(\pi\tan x)}dx\\
&=1-\frac{1}\pi\int_0^{\pi/2}\left(1+\frac{u^2-1}{1+u^2+2u\cosh(\pi\tan x)}\right)dx\\
&=\frac12+\frac{1-u^2}\pi\int_0^{\pi/2}\frac{dx}{1+u^2+2u\cosh(\pi\tan x)}
\end{align}$$
Which is as far as I could get. Perhaps we may be able to use 
$$\frac1{1+2x\cos\phi+x^2}=\sum_{n\ge0}(-1)^n\frac{\sin(n+1)\phi}{\sin\phi}x^n$$
but it sort of seems like a shot in the dark.
