Proving that $(-1)^{n+1}G_n =\int_0^\infty \frac{1}{(1+x)^n (\pi^2+\ln^2 x)} dx$ 
Prove Schroder's Integral formula: $$(-1)^{n+1}G_n =\int_0^\infty \frac{1}{(1+x)^n (\pi^2+\ln^2 x)}\mathrm dx$$ where $G_n$ are Gregory coefficients.

This is into my interest because I received an answer to this question: Integral $\int_0^1 \frac{x\ln\left(\frac{1+x}{1-x}\right)}{\left(\pi^2+\ln^2\left(\frac{1+x}{1-x}\right)\right)^2}dx$ that uses the above formula. I thought of this for some time already, but I have no idea how to start. I also searched the web, but it appears that wikipedia is the only place that it's  mentioned, of course no proof is given. I would appreciate some help! 

Edit. I have found here on AoPS a way using real methods that solves my problem.
 A: For an approach using contour integration:
By definition, $G_n$ are coefficients of $z^n$ of $z/\ln(1+z)$, it suffices to prove

For $a<1$:$$\int_0^\infty  {\frac{1}{{(x + 1 - a)({{\ln }^2}x + {\pi ^2})}}dx}  = \frac{1}{a} + \frac{1}{{\ln (1 - a)}}$$

Let $$f(z) = \frac{{{e^z}}}{{({e^z} + 1 - a)(z - \pi i)}}$$ integrate it along rectangular contour with vertices $-R, R, R+2\pi i, -R+2\pi i$, $R$ being very large. The only poles inside are $z=i\pi, \ln(1-a)+i\pi$, with residues $1/a, 1/\ln(1-a)$ respectively. Hence
$$\int_{ - \infty }^\infty  {\frac{{{e^x}}}{{({e^x} + 1 - a)(x - \pi i)}}dx}  - \int_{ - \infty  + 2\pi i}^{\infty  + 2\pi i} {\frac{{{e^x}}}{{({e^x} + 1 - a)(x - \pi i)}}dx}  = 2\pi i\left[ {\frac{1}{a} + \frac{1}{{\ln (1 - a)}}} \right]$$
Combining them (they individually diverge, but this is not an issue):
$$\int_{ - \infty }^\infty  {\frac{{{e^x}}}{{{e^x} + 1 - a}}\left[ {\frac{1}{{x - \pi i}} - \frac{1}{{x + \pi i}}} \right]dx}  = 2\pi i\left[ {\frac{1}{a} + \frac{1}{{\ln (1 - a)}}} \right]$$
the result follows via a simple substitution.
