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?

Group theory gives a pretty direct answer.  Use 
$$
G(t,x)=(\lambda t,\lambda^\beta x)\lambda_o=1
$$Then apply the group to the DEQ:
$$
t\bigg(\frac{dx}{dt}\bigg)^2+2t\frac{dx}{dt}=x\rightarrow \lambda t\bigg(\frac{\lambda ^\beta dx}{\lambda dt}\bigg)^2+2\lambda t\frac{\lambda^\beta dx}{\lambda dt}=\lambda^\beta x
$$For the group to be invariant, $2\beta-1=\beta$ or $\beta=1$.  There are two nontrivial stabilizers for this group:
$$
\mu=\frac{x}{t^\beta}=\frac{x}{t}
$$and
$$
\nu=\frac{x'}{t^{\beta -1}}=x'
$$According to Sophus Lie, your DEQ can be written in terms of the stabilizers of an invariant Lie group translation, one that preserves the structure of the smooth manifold.  Since our group is an infinite continuous group, we are able to do just that.
$$
x'^2+2x'=\frac{x}{t}\rightarrow \nu^2+2\nu=\mu
$$This is a quadratic for $\nu$.
$$
\nu=-1\pm\sqrt{1+\mu}
$$Since
$$
t\frac{d\mu}{dt}=\nu-\mu=-1\pm\sqrt{1+\mu}-\mu
$$this becomes a separable equation.
$$
\frac{d\mu}{-(1+\mu)\pm\sqrt{1+\mu}}=\frac{dx}{x}
$$This integrates and simplifies to give
$$
1\mp\sqrt{1+\mu}=\frac{C}{\sqrt{t}}
$$and after substituting for $\mu$ and simplifying again, we arrive at the solution:
$$
x=C^2-2C\sqrt{t}
$$There are other, perhaps easier ways of getting to this solution, but few people know about applied Lie Theory so I like to "beat the drum" whenever possible.
