Hung Up On Particular Differential Equation (Separable)

Question

$(e^y+1)^2e^{-y}dx+(e^x+1)^3e^{-x}dy=0$

This is a question from the back of my differential equations text book. It's not a homework question, I'm just trying to work through some of the problem types that I have trouble with, and this is one of them. So far I've got this...

re-arranging algebraically

$\frac{e^y}{(e^y+1)^2}dx+\frac{e^x}{(e^x+1)^3}dy=0$

re-arranging algebraically some more

$e^x(e^x+1)^{-3}dx=-e^y(e^y+1)^{-2}dy$

integrating both sides

$\int{e^x(e^x+1)^{-3}dx}=-\int{e^y(e^y+1)^{-2}dy}$

gives

$(e^y+1)^{-1}+2(e^x+1)^{-2}=c$

There are two issues I'm having with this solution. One is that I had to do the integrals with a CAS calculator in order to get them. I don't recall learning to do these types of integrals in Calculus 2, and two, my solution doesn't quite match that which is giving in the answer section in the back of the text book.

The book gives this as a solution

$(e^x+1)^{-2}+2(e^y+1)^{-1}=c$

The only differences is that the $y's$ and the $x's$ are in the complete opposite places, and the exponents are in opposite places. I've done the problem several times, and keep arriving at the same solution. Is my solution valid?

I would love it if someone who understands what's going on here could shed some light. I would also love to understand how to do the two integrals here by hand.

Thank You in advance

Answer 1

$$(e^y+1)^2e^{-y}dx+(e^x+1)^3e^{-x}dy=0$$ $$(e^y+1)^2e^{-y}dx=-(e^x+1)^3e^{-x}dy$$ $$\int \frac {e^{x}}{(e^x+1)^3}dx=-\int \frac {e^{y}}{(e^y+1)^2}dy$$ Substitute $u=e^x+1 \text { with }du=e^xdx$ and $v=e^y+1\text { with } dv=e^ydy$ $$\int \frac {du}{u^3}=-\int \frac {dv}{v^2}$$ $$\frac {1}{2u^2}+ \frac {1}{v}=K$$ $$\boxed{\frac {1}{(e^x+1)^2}+\frac {2}{e^y+1}=K}$$

Answer 2

To do the integral \begin{eqnarray*} \int \frac{e^{x}} {(1+e^{x})^3} dx \end{eqnarray*} let $u=1+e^x$ \begin{eqnarray*} \int \frac{du} {u^3} dx = \frac{-1}{\color{red}{2}u^2}+C = \frac{-1}{\color{red}{2}(1+e^x)^2} +C. \end{eqnarray*}