I am trying to do this problem using a somewhat similiar book example...
dy/dx = (e^y + 1)(x-2)^9
d/dy (e^y+1) = d/dx (x-2)^9
integral d/dy (e^y +1) = integral d/dx (x-2)^9
ln[y+1]^10 = 9(x-2)^8
$\displaystyle y' = (e^y + 1)(x - 2)^9$ .............divide both sides by $\displaystyle e^y + 1$
$\displaystyle \Rightarrow \frac {y'}{e^y + 1} = (x - 2)^9$ ........now split up the y' and set up the integration
$\displaystyle \Rightarrow \int \frac {1}{e^y + 1}dy = \int (x - 2)^9 dx$ ........as it stands, the left hand side is hard to integrate
$\displaystyle \Rightarrow \int \frac {e^y + 1 - e^y}{e^y + 1} dy = \int (x - 2)^9dx$ ......add $\displaystyle e^y - e^y$ (which is zero so we're not changing anything) to the numerator on the left.
$\displaystyle \Rightarrow \int 1dy - \int \frac {e^y}{e^y + 1}dy = \int (x - 2)^9dx$ ........split up the integral on the left
$\displaystyle \Rightarrow y - \ln \left|e^y + 1 \right| = \frac {1}{10}(x - 2)^{10} + C$ .....integrated, using substituion on the second integral on the left
if you have any questions say so
We have,
$\displaystyle y' = (e^y+1)(x-2)^9$
Divide,
$\displaystyle \frac{y'}{e^y+1} = (x - 2)^9$
Integrate,
$\displaystyle \int \frac{y'}{e^y+1} dx = \int (x -2)^9 dx$
Substitution rule,
$\displaystyle \int \frac{1}{e^y+1} dy = \int (x-2)^9 dx$
Now you need to find,
$\displaystyle \int \frac{1}{e^y+1} dy = \int \frac{e^y}{e^y(e^y+1)}dy$
Let $\displaystyle t=e^y \mbox{ with }t'=e^y$ thus,
$\displaystyle \int \frac{1}{t(t+1)} dt = \int \frac{1}{t} - \frac{1}{t+1} dt = \ln |t| - \ln |t+1| +C= \ln \left| \frac{t}{t+1} \right| +C = \ln \frac{e^y}{e^y+1} +C =$$\displaystyle - \ln (1+e^{-y})+C$
Thus, we find that,
$\displaystyle - \ln (1+e^{-y}) = \frac{1}{10}(x-2)^{10} +C $