Solve $\int \sqrt{7x + 4}\,dx$

Solve $\int \sqrt{7x + 4}\,dx$

I need to solve the following integral
$$\int \sqrt{7x + 4}\,dx$$
I did the following steps:
\begin{align} \text{Let} \, u &= 7x+4 \quad \text{Let} \, du = 7 \, dx \\
\int &\sqrt{u} \, du\\ &\frac{2 (7x+4)^{3/2}}{3} \end{align}
The solution is: $\frac{2 (7x+4)^{3/2}}{21}$. I am having some trouble
understanding where the denominator, $21$, comes from (is it because you
integrate $du$ also and thus, $7 dx$ becomes $\frac{1}{7}$?). I believe
this is some elementary step that I am missing. Can someone please explain
to me this?
Thanks!
P.S Is it correct to say "solve the integral"?