The integral ∫ x 2 x + 2 x x log 2 x   d x is equal to

Given,

$I=\int \left({\left(\frac{x}{2}\right)}^x+{\left(\frac{2}{x}\right)}^x\right){\log }_{2}x dx$

Now, Let ${\left(\frac{x}{2}\right)}^x=t$

$\Rightarrow x{\log }_2\left(\frac{x}{2}\right)={\log }_{2}t$

$\Rightarrow x{\log }_e\left(\frac{x}{2}\right)·{\log }_{2}e={\log }_{e}t·{\log }_{2}e$

On differentiating both sides we get :
${\log }_e\left(\frac{x}{2}\right)+x·\frac{2}{x}·\frac{1}{2}=\frac{1}{t}\frac{dt}{dx}$

$\Rightarrow {\log }_e\left(\frac{x}{2}\right)+1=\frac{1}{t}\frac{dt}{dx}$

Then solution is not possible as there is no proper substitution.

Note: This question was bonus in Jee Mains 2023 April session.