Calculus Level 5

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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