Proving that $\int_0^\infty\frac{x^2 {\rm d} x}{e^x+1} = \frac{3}{4}\int_0^\infty\frac{x^2 {\rm d}x}{e^x-1}$ without zeta functions.

Write $e^x-1=(e^{x/2}+1)(e^{x/2}-1)$ so $$\frac 2{e^x-1}=\frac1{e^{x/2}-1}-\frac1{e^{x/2}+1},$$ so $$2\int_0^\infty\frac {x^2}{e^x-1}\,dx=8\int_0^\infty\frac {(x/2)^2}{e^{x/2}-1}\frac {dx} 2- 8\int_0^\infty\frac {(x/2)^2}{e^{x/2}+1}\frac{dx}2$$ and so on.

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