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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} dx = \frac{\pi^4}{15} -\frac{\pi^4}{15}  \end{equation}  But this would then imply \begin{equation}  \frac{(e^x(x-3)+3)}{(e^x-1)}=1 \frac{(e^x(x-3)+3)}{(e^x-1)}=-1  \end{equation}  Which doesn't appear to be true...  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.  A simpler set of examples with the normal integral on the left, and the answer on the right