Yeah Y2J!

$\begin{aligned} Y & = \int_0^1 \frac{\color{#3D99F6}{\ln(1+x)}}{x} dx \\ & = \int_0^1 \frac{\color{#3D99F6}{\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n+1}x^n}{n}}}{x} dx \quad \quad \small \color{#3D99F6}{\text{Maclaurin series of }\ln(1+x)} \\ & = \int_0^1 \sum_{n=1}^\infty \frac{(-1)^{n+1}x^{n-1}}{n} dx \\ & = \sum_{n=1}^\infty \int_0^1 \frac{(-1)^{n+1}x^{n-1}}{n} dx \\ & = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^2} \\ & = \sum_{n=1}^\infty \frac{1}{n^2} - 2 \sum_{n=1}^\infty \frac{1}{(2n)^2} \\ & = \sum_{n=1}^\infty \frac{1}{n^2} - \frac{1}{2} \sum_{n=1}^\infty \frac{1}{n^2} \\ & = \frac{1}{2} \sum_{n=1}^\infty \frac{1}{n^2} \\ & = \frac{\pi^2}{12} \end{aligned}$

$\Rightarrow \left \lfloor \dfrac{24Y}{\pi^2} \right \rfloor = \left \lfloor 2 \right \rfloor = \boxed{2}$