Who's up to the challenge? 56

The given integral can be written as $I=\int_{0}^\infty\left[\frac{\ln(1+e^{2x})}{1+e^{3x}}+\frac{\ln(1+e^{-2x})}{1+e^{-3x}}\right]dx\\=\int_{0}^\infty\left[\frac{(2x+\ln(1+e^{-2x}))e^{-3x}}{1+e^{-3x}}+\frac{\ln(1+e^{-2x})}{1+e^{-3x}}dx\\=\int_0^\infty \ln(1+e^{-2x})dx+\int_0^\infty \frac{2xe^{-3x}}{1+e^{-3x}}dx\right]$ Now, $\int_0^\infty \ln(1+e^{-2x})dx=x\ln(1+e^{-2x})\left|_{0}^{\infty}\right.+\int_{0}^\infty \frac{2xe^{-2x}}{1+e^{-2x}}dx\\=\frac{1}{2}\int_{0}^{\infty}\frac{xe^{-x}}{1+e^{-x}}dx$ and $\int_{0}^\infty \frac{2xe^{-3x}}{1+e^{-3x}}dx=\frac{2}{9}\int_{0}^\infty \frac{xe^{-x}}{1+e^{-x}}dx$ Now, $\int_0^{\infty}\frac{x^ae^{-x}}{1+e^{-x}}dx\\=\sum_{k\ge 0}(-1)^k\int_{0}^\infty x^ae^{-(k+1)x}dx\\=\sum_{k\ge 0}\frac{(-1)^k}{(k+1)^{a+1}}\Gamma(a+1)\\=\Gamma(a+1)\zeta(a+1)(1-2^{-a})$ Thus, the integral is $I=\left(\frac{1}{2}+\frac{2}{9}\right)\Gamma(2)\zeta(2)(1-2^{-1})=\frac{13\pi^2}{216}$ Thus the answer is $\boxed{231}$ .