We know that \(\displaystyle \int_0^\infty \dfrac{x^2}{e^x-1}\; dx = 2\zeta(3)\). But if we also knew the value of \(\displaystyle \int_0^\infty \left(\dfrac{x}{e^x-1}\right)^2\; dx\), then what would have been the value of:

\[\int_0^\infty \dfrac{x^2}{e^x-1}\; dx + \int_0^\infty \left(\dfrac{x}{e^x-1}\right)^2\; dx = ?\]

If the answer can be expressed as \(\dfrac{\pi^A}{B}\) for positive integers \(A\) and \(B\) then find \(A\times B\).

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