deletions | additions       diff --git a/untitled.tex b/untitled.tex  index 729d0c1..a66aa23 100644  --- a/untitled.tex  +++ b/untitled.tex 
...
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}  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}  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.