Let $F(x)$ be an arbitrary continuously differentiable function, and consider the following differential equation
$$ y'+ \frac{F'(x)}{F(x)}y= \frac{1}{F(x)}.$$
If I use the Integrating Factor which says that for an equation $$ y'+ P(x)y= Q(x),$$
the general solution is
$$ y=e^{-\int P(x)dx}\int Q(x)e^{\int P(x)dx}dx+Ce^{-\int P(x)dx}.$$
Using the fact that
$$\int \frac{F'(x)}{F(x)} dx = \ln(F(x)), $$
this simplifies to
$$ y=(1+C)F(x)^{-1}.$$
However, if I substitute this solution into the initial differential equation I get
$$ -\frac{(1+C)F'(x)}{F(x)^2} + \frac{F'(x)}{F(x)}\frac{(1+C)}{F(x)} = \frac{1}{F(x)}, $$
or
$$ 0 = \frac{1}{F(x)}. $$
Where is my mistake?
You don't really need an integrating factor: $$y'+ \frac{F'(x)}{F(x)}y= \frac{1}{F(x)}$$ $$F(x)y'+ F'(x)y= 1 $$ $$(F(x)y)'= 1$$ $$F(x)y= x+c$$ $$ \implies y(x)= \dfrac {\color{blue}{x+c}}{F(x)}$$ And not $y(x)= \dfrac {\color{red}{1+c}}{F(x)}$
$$y=e^{-\int P(x)dx}\int Q(x)e^{\int P(x)dx}dx+Ce^{-\int P(x)dx}.$$ $$y=\dfrac {1}{F(x)}\int dx+\dfrac C{F(x)}$$ $$y=\dfrac {x}{F(x)}+\dfrac C{F(x)}$$