Proving that $(-1)^{n+1}G_n =\int_0^\infty \frac{1}{(1+x)^n (\pi^2+\ln^2 x)} dx$

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.