Show that there is a maximal solution and determine it Consider the IVP
$$
x'=\frac{1}{1+\lvert x\rvert},~~x(4)=2.
$$
(1) Show that there exists a maximal solution.
(2) Determine the maximal solution.

(1) I guess that $f\colon\mathbb{R}\to\mathbb{R}^+, x\mapsto \frac{1}{1+\lvert x\rvert}$ is the right-hand side of the IVP, isn't it? It is continuous in $x$ and continuously differentiable in $x$, hence for each initial pair $(t_0,x_0)$ there exists an unique maximal solution, in particular for $t_0=4, x_0=2$. Is this okay?
(2) Do not know how to determine the maximal solution now. Maybe by separation of the variables?
Then I get
$$
x+\frac{x\cdot \lvert x\rvert}{2}=t+c,
$$
where c is a constant.
Do not know how to continue.
 A: Separation of the variables is the thing you should do. Remember that the solutions of autonomous differential equation are always monotone in special here this is trivial as your derivative is always strictly positive. So there is a neighbourhood of $4$ such that $x(t)$ will be strictly positive which simplifies your equation. 
Now just solve for $x$ and you are done. 
As I said in a neighbourhood of $4$ we have the equation 
$$ \frac{x^2}{2}+x=t+c$$ and with the initial value we see that $c=0$. 
This is equivalent to 
$$ \frac{1}{2} (x^2+2x+1-1-2t)=\frac{1}{2}((x+1)^2-1-2t)=0$$ 
But that means that $x(t)$ must be given by 
$$x(t)= \pm \sqrt{2t+1} -1$$ 
and the initial value tells us that it is $x(t)= \sqrt{2t+1}-1$ in a neighbourhood of $4$. Indeed we can chose the neighbourhood to be $(0,\infty)$ because with our simplification we assumed $x$ to be non negative. As we don't have a blow up in $0$ one might look at the initial value problem 
$$x'(t)=\frac{1}{1+|x|} $$ with $x(0)=0$. As this is an autonomous differential equation we know that $x(t)<0$ for $t<0$, hence we have now 
$$-\frac{x^2}{2} + x=t+c$$ which gives us 
$$x(t)=1-\sqrt{1-2t}$$ as a solution for $t<0$. 
A: Your application of the separation formula is correct. For positive $x$ you get
$$
2x+x^2=2t+2c\implies x=\sqrt{1+2t+2c}-1=\frac{2t+2c}{\sqrt{1+2t+2c}+1}\text{ for } t+c\ge 0
$$
and for the arcs of negative values of $x$
$$
2x-x^2=2t+2c\implies x=1-\sqrt{1-2t-2c}=\frac{2t+2c}{1+\sqrt{1-2t-2c}}\text{ for } t+c<0
$$
which can be unified to the globally valid formula
$$
x(t)=\frac{2t+2c}{1+\sqrt{1+2|t+c|}}
$$
In the arc containing the initial point $x(4)=2$ one gets $c=0$, thus the global solution is
$$
x(t)=\frac{2t}{1+\sqrt{1+2|t|}}
$$
