Just for my own understanding of how exactly integration works, are these steps correct:
$$\begin{align}\int x\,d(x^2) \qquad &\implies x^2 = u \\ & \implies x= \sqrt{u}\end{align}$$
Thus, it becomes $$\int\sqrt{u}\,du = \frac{2}{3}u^{3/2} \implies {2\over3}x^3$$
Yes, except for an integration on the LHS of the last line. Also make sure to keep track of your limits. You would get the same expression when you use $d(x^2) = 2x dx$
While fundamentally correct, aside from your lack of a $+C$ constant, notationally your work is a bit of a mess (before someone edited it for clarity's sake).
First, we start with
$$\int x d(x^2)$$
From here, we make the substitution $u = x^2$. This gives $x = \sqrt{u}$. Thus,
$$x d(x^2) = \sqrt{u} du \;\;\; \Rightarrow \;\;\; \int xd(x^2) = \int u^{1/2}du = \frac{2}{3}u^{3/2} + C$$
Reutilizing our substitution, we finally get our answer:
$$\frac{2}{3}u^{3/2} + C = \frac{2}{3}x^3 +C$$
Your errors make it hard to follow your thought process, because $\sqrt{u} du \neq \frac{2}{3}u^{3/2}$, when you assert they're equal in your last line.
As a note, this question was asked before: you might want to check more thoroughly next time. The original question.
