Well you are right in making x the subject of the equation. This is because you cannot integrate a logarithm, but you can integrate a function that involves e.
So you should get x = e^y + 2 as your new equation.
When you integrate this equation, you are integrating with respect to y, hence why the limits have to be changed to those on the y axis instead.
Integrating, you get e^y + 2y with the limits on In5 and 0.
Remember that this finds the area between the curve and the y axis. (This is important later).
When you substitute In5 and 0 into the integral, you get:
(5 + 2In5) - (1) = 4 + 2In5
We need the area between the x axis and the curve, not the y axis and the curve, so consider the graph as a rectangle with a length of 7 and a height of In5. This means the TOTAL area is 7In5.
This means that the area under the curve is 7In5 - (4 + 2In5) = 5In5 - 4
I did that pretty quickly - sorry if there are mistakes :S
I hope it helps nonetheless.
=]
