Divide $ yy' = (y + 1)^2 $ by $ (y + 1)^2 $ I was reading from Ordinary Differential Equations (Lesson 4C page 36) and came across this question:
Find a 1-parameter family of solutions of the differential equation
$$ (a) \quad yy' = (y + 1)^2 $$
and the particular solution for which $\quad y(2) = 0$
The first step was to divide $ (a) $ by $ (y + 1)^2 $
and they had 
$$ \int{\frac{y}{(y + 1)^2} dy } = \int dx \quad,\quad y \neq -1$$
but when I was trying to divide $ (a) $ by $ (y + 1)^2 $
I had 
$$ \frac{yy'}{(y+1)^2} = \frac{(y+1)^2}{(y+1)^2} $$
$$ => \quad \frac{yy'}{(y+1)^2} = 1$$ 
 A: Just for your curiosity.
Starting from @Axion004's answer
$$\frac{1}{y+1}+\ln|y+1|=x+C$$ the condition $y(2)=0$ leads to $C=-1$. Now, let $z=\frac{1}{y+1}$ to get
$$z-\log(|z|)=x-1\implies z=-W\left(-e^{1-x}\right)$$ where appears the  branches of Lambert function (your will learn sooner or later about this beautiful function). Then
$$y=-\frac{1}{W_{0}\left(-e^{1-x}\right)}-1\qquad \text{and} \qquad y=-\frac{1}{W_{-1}\left(-e^{1-x}\right)}-1$$
A: They are doing the same thing as you are, but they have just gone a little bit further.
Note that $y'=\mathrm dy/\mathrm dx$, so (assuming $y\neq-1$) we may separate
$$\frac{y}{(y+1)^2}\,\mathrm dy=\mathrm dx,$$
then integrate
$$\int\frac{y}{(y+1)^2}\,\mathrm dy=\int\mathrm dx.$$
A: In differential equations and calculus it is often assumed that
$$y'=\frac{dy}{dx}$$
where $y$ is the dependent variable and $x$ is the independent variable. In your case, given $y \neq 1$
$$\frac{yy'}{(y+1)^2} = \frac{(y+1)^2}{(y+1)^2}=1 \implies \frac{y}{(y+1)^2}\frac{dy}{dx}=1 \implies \frac{y}{(y+1)^2}\,dy=dx$$
by which you can integrate
$$\int \frac{y}{(y+1)^2}\,dy=\int dx$$
by observing that the left-hand side has the following partial fraction decomposition
$$\frac{y}{(y+1)^2}=\frac{1}{y+1}-\frac{1}{(y+1)^2}$$
so that
$$\int \frac{y}{(y+1)^2}\,dy=\int \left(\frac{1}{y+1}-\frac{1}{(y+1)^2}\right)dy=\frac{1}{y+1}+\ln|y+1|+C$$
therefore the implicit solution is 
$$\frac{1}{y+1}+\ln|y+1|=x+C$$
