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"?

When you made the u-substitution, you took $u=7x+4$ and hence $du=7 dx$. You forgot this factor of 7! In particular, $dx=du/7$.
It helps to write out the $dx$ in the integral:
$$\int \sqrt{7x+4} dx=\int \frac{\sqrt{u}}{7} du.$$
