Given,
I=∫((2x)x+(x2)x)log2xdx
Now, Let (2x)x=t
⇒xlog2(2x)=log2t
⇒xloge(2x)⋅log2e=loget⋅log2e
On differentiating both sides we get :
loge(2x)+x⋅x2⋅21=t1dxdt
⇒loge(2x)+1=t1dxdt
Then solution is not possible as there is no proper substitution.
Note: This question was bonus in Jee Mains 2023 April session.