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We could differentiate the integrand and times by x, then flip the sign and gain the result \begin{equation}  \int_0^\infty \frac{e^x(x-3)+3)x^3}{e^x-1)^2} \frac{e^x(x-3)+3)x^3}{(e^x-1)^2}  dx = \frac{\pi^4}{15} \end{equation}  In theory for any function which ends up as a polynomial, the differentiation and multiplication with $x$ should leave the form unchanged except for a numeric factor, which will drop out in some circumstances.