Book tells me to solve separable diffeq with integrating factor, am I missing something? Maybe this is for purely pedagogical purposes but the book I am using instructs me to solve
$$
ydx+(1-x)dy=0
$$
By finding an appropriate integrating factor and solving. But 
$$
ydx+(1-x)dy=0\Rightarrow (1-x)dy=-ydx\Rightarrow \frac{-dy}{y}=\frac{dx}{1-x}\\
\Rightarrow -\ln(y)=-\ln|1-x|+c \Rightarrow y(x)=A(1-x)
$$
Which gives an explicit solution for $y(x)$.
 A: As suggested, I did it the books way as well, but it didn't quite work out. I thought about posting another question, but thought it would be more useful for future users to have it all in one thread. Here is my attempt at the answer.
$$ydx+(1-x)dy=0$$
Yields an integrating $I(x)$ factor of
$$\frac{1}{1-x}(1-(-1))=\frac{2}{1-x}\Rightarrow \exp(2\int\frac{dx}{1-x})\\
=\exp(-2\ln|1-x|)\Rightarrow I(x)=\frac{1}{(1-x)^2}$$
Which yields the exact equation
$$
\frac{y}{(1-x)^2}dx+\frac{1}{1-x}dy=0
$$
And solving for some $G(x,y)=c$
$$
\frac{\partial G}{\partial y}=\frac{1}{1-x}\Rightarrow G(x,y)=\ln|1-x|+h(x)
$$
and
$$
\frac{\partial G}{\partial x}=\frac{y}{(1-x)^2}=\frac{1}{1-x}+h'(x)\\
h'(x)=\frac{y+x-1}{(1-x)^2}\Rightarrow h(x)=y\int\frac{dx}{(1-x)^2}+\int\frac{(x-1)dx}{(1-x)^2}\\
=y\int\frac{dx}{(1-x)^2}-\int\frac{dx}{(1-x)}\Rightarrow h(x)=\frac{y
}{1-x}-\ln|x|+c\\
$$
Yielding the class of solutions:
$$
G(x,y)=c=\ln|1-x|-\ln|x|+\frac{y
}{1-x}\Rightarrow [c-\ln|\frac{x-1}{x}|](x-1)=y(x)
$$
