I have some questions to ask if that's okay 
1. For In (x) = In (2e)
can you drop 'In' because it is on both sides? Does this work for logs as well? ( I faintly remember it does?)
2. why do you ant differentiate In x? (isn't it the gradient of y = x In x -x )
3. I was wondering don't you have to have the integral of y = 1 + In 2 minus the integral of In x to find the shaded area?
1. Yes, it's saying if you take the log of x it's the same as taking the log of 2e. Therefore, x must be 2e. Alternatively, you could solve it saying x=e
ln(2e)=2e
2. You want to find the integral of ln(x) (area under the graph). Since you know that
= \ln x, \text{ it follows that } x \ln x -x = \int \ln x dx )
which is what you want to find

derivative is like the opposite of integration
3. The integral of ln x is the area under the graph (the shaded part). If you wanted to find the white part under the graph of y = 1 + In(2), you could use
 dx - \int \ln x dx)
(with bounds) but that isn't required here.
Hope this helps

Edited to put in latex code

Also never be afraid of asking questions!