Solve $\frac{2 f'}{(f-1)^2}=1$ with initial condition $f(0)=c$ I'm trying to solve this by writing it as the derivative of a log of a polynomial but I can't make it work. Any hints?
 A: Consider the function $f$ is a function of $x$.
Now $\frac{2 f'}{(f-1)^2}=1\implies \frac{df}{(f-1)^2}=\frac{1}{2} dx\implies -\frac{1}{f-1}=\frac{1}{2}x + a$, where $a$ is integrating constant.
Given that $f(0)=c$, then 
$ -\frac{1}{c-1}=0 + a\implies a=-\frac{1}{c-1} $
hence the solution of the given differential equation is $$-\frac{1}{f-1}=\frac{1}{2}x  -\frac{1}{c-1}\implies\frac{1}{f-1}=-\frac{1}{2}x +\frac{1}{c-1} $$ 
A: With
$g = f - 1, \; g(0) = f(0) - 1 = c - 1, \tag 1$
we have
$g' = f'; \tag 2$'
thus,
$\dfrac{2f'}{(f - 1)^2} = \dfrac{2g'}{g^2} = 1, \tag 3$
whence
$g^{-2}g' = \dfrac{1}{2}, \tag 4$
or
$(-g^{-1})' = \dfrac{1}{2}; \tag 5$
we integrate 'twixt $0$ and $x$:
$-g^{-1}(x) - (-g^{-1}(0)) = \displaystyle \int_0^x (-g^{-1}(s))\; ds = \int_0^x \dfrac{1}{2} \; ds = \dfrac{1}{2}x, \tag 6$
and perform a few simple algebraic maneuvers using
$g^{-1}(0) = (c - 1)^{-1}: \tag 7$
$-g^{-1}(x) + g^{-1}(0) = \dfrac{1}{2}x; \tag 8$
$-g^{-1}(x) + (c - 1)^{-1} = \dfrac{1}{2}x; \tag 9$
$g^{-1}(x) = (c - 1)^{-1} - \dfrac{1}{2}x = \dfrac{1}{c - 1} - \dfrac{1}{2}x$ $= \dfrac{2}{2(c - 1)} - \dfrac{(c - 1)x}{2(c - 1)} = \dfrac{2 - (c - 1)x}{2(c - 1)}; \tag{10}$
$g(x) = \dfrac{2(c - 1)}{2 - (c - 1)x} = 2(c - 1)(2 - (c - 1)x)^{-1}. \tag{11}$
We Check:
Differentiating (11) yields
$g'(x) = -2(c - 1)(2 - (c - 1)x)^{-2} (-(c - 1))$
$= 2(c - 1)^2(2 - (c - 1)x)^{-2} = 2(((c - 1)(2 - (c - 1)x)^{-1})^2 = \dfrac{g^2}{2}, \tag{12}$
in accord with (3).  It is also a simple matter to verify that
$g(0) = c - 1. \tag{13}$
Finally, since $g(x)$ is the solution to (4), 
$f(x) = g(x) + 1 = 2(c - 1)(2 - (c - 1)x)^{-1} + 1 \tag{14}$ 
satisfies (3) with
$f(0) = g(0) + 1 = c. \tag{15}$
