\[\large \int_0^1 \dfrac{(\ln x)^2}{1-x} \, dx = 2a \int_1^\infty \dfrac{\lfloor x \rfloor }{x^{a+1}} \, dx \]

Given that \(a\) is a positive integer constant satisfying the equation above, find \(a\).

**Bonus**: Prove that \( \displaystyle a \int_1^\infty \dfrac{\lfloor x \rfloor }{x^{a+1}} \, dx = \zeta(a) \).

**Notation**: \( \zeta(\cdot) \) denotes the Riemann zeta function.

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