Let for $n>1,n\in N.I_{2n}=\underset{2001}{\overset{4002}{\int }}\frac{1+{(-1)}^{n}2{\log }_{x}2}{{(1+{\log }_{x}2)}^2}dx$ and $I_2=\underset{2001}{\overset{4002}{\int }}{\left({\log }_{2x}\sqrt{2}\right)}^{2}dx$ Where $x\in (1,\infty )$ then which of the following is/are TRUE

$I_2=\frac{1}{4}\underset{2001}{\overset{4002}{\int }}{\left({\log }_{2x}2\right)}^{2}dx\Rightarrow 4I_2=\underset{2001}{\overset{4002}{\int }}{\left({\log }_{2x}2\right)}^{2}dx$

$4I_2+I_{4n}=\underset{2001}{\overset{4002}{\int }}\{\frac{1}{{(1+{\log }_{2}x)}^2}+\frac{{\left({\log }_{2}x\right)}^2+2{\log }_{2}x}{{(1+{\log }_{2}x)}^2}\}dx=\underset{2001}{\overset{4002}{\int }}\frac{(1+2{\log }_{2}x+{\left({\log }_{2}x\right)}^2)}{{(1+{\log }_{2}x)}^2}dx$

$\underset{2001}{\overset{4002}{\int }}1dx=4002-2001=2001$

$4I_2+I_{4n+1}=\underset{2001}{\overset{4002}{\int }}(\frac{1}{{(1+{\log }_{2}x)}^2}+\frac{{\left({\log }_{2}x\right)}^2-2{\log }_{2}x}{{(1+{\log }_{2}x)}^2})dx$

$$\begin{gathered} =\underset{2001}{\overset{4002}{\int }}\frac{{(l{\mathrm{og}}_{2}x-1)}^2}{{(1+{\log }_{2}x)}^2}dx=\underset{2001}{\overset{4002}{\int }}{\left(\frac{{\log }_2\frac{x}{2}}{{\log }_{2}2x}\right)}^{2}dx \\ =\underset{2001}{\overset{4002}{\int }}{\left({\log }_{2x}\frac{x}{2}\right)}^{2}dx \end{gathered}$$

Put $2x=t\Rightarrow dx=\frac{dt}{2}\Rightarrow \frac{1}{2}\underset{4002}{\overset{8004}{\int }}{\left(lo{g}_t\frac{t}{4}\right)}^2.dt$