First order ODE: $tx'(x'+2)=x$ $$tx'(x'+2)=x$$
First I multiplied it:
$$t(x')^2+2tx'=x$$
Then differentiated both sides:
$$(x')^2+2tx'x''+2tx''+x'=0$$
substituted $p=x'$ and rewrote it as a multiplication
$$(2p't+p)(p+1)=0$$
So either $(2p't+p)=0$ or $p+1=0$
The first one gives $p=\frac{C}{\sqrt{T}}$
The second one gives $p=-1$. My question is how do I take the antidervative of this in order to get the answer for the actual equation?

Your differential equation is a complicated one. I give some hints on $I=]0,+\infty[$ (the study has to be done also on $]-\infty,0[$).
A) First note that on $I$, your equation is $\displaystyle (x^{\prime}(t)+1)^2=\frac{x(t)+t}{t}$. We see that $x_0(t)=-t$ is a solution. For any solution, we  must have $x(t)+t\geq 0$. 
Now suppose that on $I$, we have $x(t)+t >0$ for all $t$. Then $x^{\prime}+1\not =0$ for all $t$; by Darboux's theorem, $x^{\prime}+1$ has a constant sign on $I$. With $\varepsilon\in \{\pm 1\}$, we can write $\displaystyle \frac{x^{\prime}+1}{\sqrt{x(t)+t}}=\frac{\varepsilon}{\sqrt{t}}$, and hence $2\sqrt{x(t)+t}=2\varepsilon\sqrt{t}+C$ where $C$ is a constant. Now $x(t)=C^2/4+C\varepsilon \sqrt{t}$. Of course, we have to verify that all these solutions are really solutions(and verify what conditions on $C$ and $\varepsilon$ gives $x(t)+t>0$ for all $t$).
B) To see how the hypothesis $x(t)+t>0$ is important, let $t_0>0$. Put $x(t)=-t$ for $0<t\leq t_0$, and $x(t)=t_0-2\sqrt{t_0t}$ if $t>t_0$. I leave to you the verification that $x(t)$ is a solution, (first show that the derivative at $t_0$ exists and is equal to $-1$). But this solution is not one of the form $C^2/4+C\varepsilon \sqrt{t}$ we have found, as they cannot agree on $]0,t_0[$.  
