Integral (square root function multiplied by exponential function) did i do it right? I'm trying to determine $\int x^3\sqrt{x^2 +1}\, dx$

I said that $u(x) = x^2 + 1$

and then that $dx = 2x\,dx$
so I rewrote the integral as 

$$\int x^3\sqrt{x^2 +1}\,2x\,dx$$

which is also 

$$\int2x^4\sqrt{x^2 +1} \,dx$$

and then it is easy to integrate, is that all legal to do?
I'm afraid not. You should not have an equation with a "differential factor" on one side and not on the other (that is, $dx=2x$ is nonsense). For more on how to deal with differential factors, you might find the second part of this answer (from "Now, if I wrote..." through "...let's get back to your problem.") useful.

What you can say is that $$\frac{du}{dx}=2x,$$ so that $$du=2x\,dx,$$ so that $$x\,dx=\frac12\,du.$$ Then your substitution gives you $$\int x^3\sqrt{x^2+1}\,dx=\frac12\int x^2\sqrt{u}\,du.$$ We're not quite there, yet, though. Can you rewrite $x^2$ in terms of $u$?
HINT:  when completing a $u$ substitution, the typical flow is as follows:

$$\int x^3\sqrt{x^2+1}dx$$

Substitute $u=x^2+1$, then $du=2xdx$ and we have $x^2=u-1$:

$$\int {(u-1)\sqrt u du\over 2}$$

From there, the remaining steps are to integrate, then reverse-substitute to obtain the function relative to $x$, and to add the unknown constant.