Little integral $\int_{0}^{\infty}\ln\Big(\frac{x^2-2x+1}{x^2+2x+1}\Big)e^{-x}dx$ related to the Gompertz constant It's a little result related to the Gompertz constant 
We have :

$$\int_{0}^{\infty}\ln\Big(\frac{x^2-2x+1}{x^2+2x+1}\Big)e^{-x}dx=-2G-2\frac{\operatorname{Ei(1)}}{e}$$
  Where $G$ is the Gompertz constant and $\operatorname{Ei}$ is the exponential integral .

This integral have an simple antiderivative .But i look for a trick to solve this .
If you have nice ideas...
Thanks a lot for your contributions !
 A: It's rather simple to manage this: For the undefined integration, just notice that
$$\log\left(\frac{x^2 - 2x + 1}{x^2+2x+1}\right) = \log\left(\left(\frac{x-1}{x+1}\right)^2\right) = 2\log\bigg|\frac{x-1}{x+1}\bigg|$$
This being said, proceed as follows:
$\bullet$ First substitution: $x+1 = t ~~~~~~~ \text{d}x = \text{d}t$, your integral becomes
$$2e\int\ln\bigg|1 - \frac{2}{t}\bigg|e^{-t}\ \text{d}t$$
$\bullet$ Second substitution: $1 - \frac{2}{t} = z ~~~~~ \text{d}t = \frac{-2}{(1-z)^2}\text{d}z$, your integral becomes
$$2e\int \ln|z| e^{-\frac{2}{1-z}} \left(\frac{-2}{(1-z)^2}\right)\ \text{d}z$$
Now just notice that the last integral is nothing but
$$2e\int \ln|z| \frac{\text{d}}{\text{d}z} \left(-e^{-\frac{2}{1-z}}\right)\ \text{d}z$$
So you can integrate by parts easily.
Remember, from the theory of Special Functions, that 
$$\int \frac{e^{-\frac{2}{1-z}}}{z}\ \text{d}z = \frac{\text{Ei}\left(2+\frac{2}{z-1}\right)}{e^2}-\text{Ei}\left(\frac{2}{z-1}\right)$$
With extrema
$\bullet$ During the first substitution, your extrema become $[0, +\infty) \to [1, +\infty)$
$\bullet$ During the second substitution, extrema become $[1, +\infty) \to [3, 1]$
I believe you can proceed alone from here.
