One major error: the integral of $\displaystyle \frac{dp}{2p}$ is NOT $\displaystyle ln(2p)$ as you can see by differentiating. Instead write it as $\displaystyle \frac{1}{2}\frac{dp}{p}$. Integrating gives $\displaystyle \frac{1}{2}ln|p|= ln|p^{1/2}|+ C$.
Personally, from $\displaystyle \frac{dp}{dx}= \frac{2p}{x}$, I would have left the "2" where it is: $\displaystyle \frac{dp}{p}= 2\frac{dx}{x}$. Integrating both sides, $\displaystyle ln|p|= 2ln|x|+ C$ so that $\displaystyle p= cx^2$ (I have dropped the absolute values because "c" can be positive or negative itself).
We cannot solve your homework problems for you. I realize you have been showing your work and all (that's a good thing!), but we cannot allow this in general. Please do not do this again.
For anyone helping please only give broad help, not specifics.
-Dan