Integrating $\int (\ln(x))^2 \, dx$

Nice work. You've got the correct answer! Just remember is showing your work to include $\,dx$ at the close of an integrals, and putting $\,dv = 1$ (it should be $\,dv = \,dx \implies v = x)$.

I'd simply suggest expressing the result of the integration by parts as follows: $$x(\ln x)^2-2[x\ln(x)-x] + C$$ We might even want to factor out $x$ from each term:

$$x\Big((\ln x)^2 - 2\ln x + 2\Big) + C$$

But your work and result are indeed correct.