How do I solve this calculus (differential equation) problem

So this problem was in my textbook, and I asked my professor about it along with quite a few others who were as entirely perplexed as I was.
The problem is as follows:
Find the general solution of the given differential equation.
dy/dx = ((e^y) + 1)(x - 2)^9

If you try u substitution, it doesn't really get you anywhere....and neither does integration by parts.
If you put the y's on one side and the x's on the other you end up with:
dy/((e^y) + 1) = (x - 2)^9 dx
but I don't see that working out well either. The x side can be easily integrated but the other doesn't work because the numerator is just dy.

Any help would be much appreciated!

dy/dx = [ (e^y + 1)(x - 2)^9 ]

separating variables

dy / (e^y + 1) = (x - 2)^9 dx

multiply top and bottom of LHS with e^y

e^y dy / e^y (e^y + 1) = (x - 2)^9 dx

let e^y + 1 = u ==> e^y = u - 1
e^y dy = du

du /u(u - 1) = (x - 2)^9 dx

=> [ 1/( u - 1) - 1/u ] du = (x - 2)^9 dx

integrate both sides

ln I u - 1I - ln I u I = (1/10)(x - 2)^10 + c

ln [(u-1)/u ] = (1/10)(x - 2)^10 + c

( u - 1) / u = C e^[1/10(x - 2)^10 ]

back substitute u = e^y + 1

e^y /(e^y + 1) = C e^[1/10(x - 2)^10 ]

I continue from :
∫ dy/((e^y) + 1) = ∫(x - 2)^9 dx
∫ e^ydy/((e^2y) + e^y) = ∫(x - 2)^9 dx
∫ d(e^y)/((e^2y) + e^y) = ∫(x - 2)^9 dx
- ∫ d((e^y)+1)/((e^y) + 1) + ∫ d(e^y)/(e^y) = ∫(x - 2)^9 dx
- Ln((e^y)+1) + Ln(e^y) = (1/10)(x-2)^10+C
Ln((e^y)/((e^y)+1)) = (1/10)(x-2)^10+C
(e^y)/(e^y)+1) = e^[(1/10)(x-2)^10+C] / 1
(e^y) = e^[(1/10)(x-2)^10+C] /[ 1 - e^[(1/10)(x-2)^10+C]]
y = Ln e^[(1/10)(x-2)^10+C] /[ 1 - e^[(1/10)(x-2)^10+C]]