Find the solution to $y' = \frac{1+y}{x^2+x}$ that satisfies $y(1) = -1$ I'm having trouble when conditions like domain on variables are present in ODE's.  
I am asked to find the solution to $y' = \dfrac{1+y}{x^2+x}$ that satisfies
$(a) \quad y(-2) = 1$
$(b) \quad y(1) = -1$
$(c) \quad y(1) = -2$
$(d) \quad y(-\frac{1}{2}) = 0$
and to state for which $x$ the solution exist.
In particular I have a problem with $(b)$.  
I reasoned like this:  
To begin with, $\dfrac{1+y}{x^2+x} = \dfrac{1+y}{x(x+1)}$ so $x \neq 0, -1$ 
If we assume $y \neq -1$ we can rewrite the equation  
$y' = \dfrac{1+y}{x^2+x} "\iff" \dfrac{dy}{1+y} = \dfrac{dx}{x(x+1)} "\iff" \int \dfrac{1}{1+y} dy = \int \dfrac{1}{x(x+1)} dx, \quad (1.)$ 
I write $"\iff"$ because that how my teacher writes, since we haven't really defined what it means to multiply or divide with $dy, dx$.  
Since $\int \dfrac{1}{x(x+1)} dx = \int \dfrac{1}{x} - \dfrac{1}{x+1} dx = C + \ln(|x|) - \ln(|x+1|) = C + \ln\left(\dfrac{|x|}{|x+1|} \right), C\in\mathbb{R} $
$\int \dfrac{1}{1+y} dy = C + \ln(|1+y|), C\in\mathbb{R}$ 
It follows from $(1.)$ that
$\ln(|1+y|) = C + \ln\left(\dfrac{|x|}{|x+1|} \right), \quad C\in\mathbb{R} \iff $
$\exp(\ln|1+y|) = \exp\left(C + \ln\left(\dfrac{|x|}{|x+1|} \right)\right), \quad C\in\mathbb{R} \iff$
$|1+y| = e^C \dfrac{|x|}{|x+1|} = C_2 \dfrac{|x|}{|x+1|}, \quad C_2 > 0 \quad (2.)$ 
Now we consider different cases based on the value of $y$ 
Case 1 : If $y > -1$ then it follows from $(2.)$ that $y =  C_2 \dfrac{|x|}{|x+1|} - 1, \quad C_2 > 0$
Case 2 : If $y < -1$ then it follows from $(2.)$ that $y = -C_2 \dfrac{|x|}{|x+1|} - 1, \quad C_2 > 0$ 
Next we consider different cases based on the value of $y$ and $x$ 
Case 1.1 : $y > -1$ and $x > 0$ implies that
$y =  C_2 \dfrac{|x|}{|x+1|} - 1 = C_2 \dfrac{x}{x+1} - 1 = \dfrac{C_2x - x - 1}{x+1} = \dfrac{(C_2-1)x - 1}{x+1} = \\ \dfrac{C_3x - 1}{x+1}, \quad C_3 > -1$ 
Case 1.2 $y > -1$ and $-1 < x < 0$ implies that
$y =  C_2 \dfrac{|x|}{|x+1|} - 1 = C_2 \dfrac{-x}{x+1} - 1 = \dfrac{-C_2x - x - 1}{x+1} = \dfrac{-(C_2+1)x - 1}{x+1} = \\ \dfrac{C_4x - 1}{x+1}, \quad C_4 < -1$ 
Similar reasoning shows that for the rest of the cases the following is true:  
Case 1.3 $y > -1$ and $x < -1$ implies that $y = \dfrac{C_3x - 1}{x+1}, \quad C_3 > -1$ 
Case 2.1 $y < -1$ and $x > 0$ implies that $y = \dfrac{C_4x - 1}{x+1}, \quad C_4 < -1$ 
Case 2.2 $y < -1$ and $-1 < x < 0$ implies that $y = \dfrac{C_3x - 1}{x+1}, \quad C_3 > -1$ 
Case 2.3 $y < -1$ and $x < -1$ implies that $y = \dfrac{C_4x - 1}{x+1}, \quad C_4 < -1$ 
Now, to solve $(a)$ we note that in this case $x = -2 < -1$ and $y = 1 > -1$ so this fits into case 1.3 where $y = \dfrac{C_3x - 1}{x+1}, \quad C_3 > -1$. Plugging in $x = -2$ and $y = 1$ we get $C_3 = 0 > -1$ so that $y = -\dfrac{1}{x+1}, \quad x < -1$ 
I reasoned similarly in $(c)$ and $(d)$ and the answers seem to be correct. 
However, in $(b)$, $y = -1$ which I assumed to be false when revriting the equation to $(1.)$. So I figured I have to try a different way than using any of the derived formulas for $y$. Just by inspecting the original equation $y' = \dfrac{1+y}{x^2+x}$ I noticed that $y(x) = -1$ solves the equation for all $x \neq 0, -1$, but in the answer they state that $y(x) = -1, \quad x > 0$ is the correct solution. So I wonder, why is the condition $x > 0$ necessary? Couldn't it just as well be $x < -1$ or $-1 < x < 0$ ?

If you consider $x < 0$ too, there will not be a unique solution to the IVP. Indeed, any solution to the IVP on the three intervals $(-\infty, -1), (-1, 0), (0, \infty)$ can be spliced together to form a solution to the differential equation. So long as the function is constantly $-1$ on $(0, \infty)$ (and whatever other solution on the other two intervals) it will satisfy the IVP.
For example, the following satisfies the IVP:
$$y(x) = \begin{cases} \frac{-3x}{x + 1} - 1 & \text{if } x < -1 \\ \frac{2x}{x + 1} - 1 & \text{if } -1 < x < 0 \\ -1 & \text{if } x > 0. \end{cases}$$
The question seems to want the unique solution locally around $x = 1$ and the largest domain on which it's unique. The question doesn't sound like it makes this clear.
