Ah there we go, searching back through the original C3 thread got the right equations for 9..
y=ln|2x/x+1|
Here we've mixed up question 8 and question 9. the one where dy/dx = -0.5 and is therefore not symmetrical across y=x is the question 8 curve.
On this question we had to show that the 2nd line on the graph (there was a graph with 2 lines on) was the inverse of the line labelled y = ln|2x/x+1|. The line was labelled as y = (e^x)/(2 - e^x) so it was a show that question.
y = ln|2x/x+1|
inverse: x = ln|2y/y+1|
e^x = 2y/(y+1)
(y+1)e^x = 2y
ye^x + e^x = 2y
e^x = 2y - ye^x
e^x = y(2 - e^x)
(e^x)/(2 - e^x) = y as required
Another part of the question was showing that the integral of (e^x)/(2 - e^x) between x=0 and x=ln|4/3| is equal to ln|3/2|
let u = (2 - e^x)
du/dx = -e^x
new limits: when x=ln|4/3|, u = 2/3, when x=0, u=1
I = - integral(1/u) du between u=1 and u=2/3
= [-lnu] between 1 and 2/3
= -ln(2/3) - -ln(1)
=-ln|2/3|
=ln|3/2| as required
Then finally it asked you to find the shaded area on the graph (under y=ln|2x/x+1|) and show that it is equal to ln|32/27| (between the point the graph crosses the x axis and x=2) = between x=1 and x=2
To do this the easiest way is that because the other line is the inverse of the first, you can find the area of the equivalent part of the curve y = (e^x)/(2-e^x), which is between x=0 and x=ln|4/3|.
Therefore the area of the shaded region is the area of the rectangle enclosed by y=0, y=2, x=0 and x=ln|4/3| and then taking away the area that you have worked out in the previous question
=2ln|4/3| - ln|3/2|
=ln|16/9| - ln|3/2|
=ln16 - ln9 - ln3 + ln2
=ln|(16*2)/(9*3)|
=ln|32/27| as required
now time for some dinner :L