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I=∫01(2x−1)lnx1−x2 dx\large I = \int_0^1 \dfrac{(2x-1)\ln x}{1-x^2} \, dxI=∫011−x2(2x−1)lnxdx
Find the value of the closed form of the above integral III. Submit your answer as ⌊24I⌋\lfloor 24I \rfloor⌊24I⌋.
Notation: ⌊⋅⌋ \lfloor \cdot \rfloor ⌊⋅⌋ denotes the floor function.
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