Is this substitution answer correct? I'm learning more about how to solve integrals with substitution.  Before, I was relying on the formula
$\int f(g(x))g'(x)dx = \int f(u)du$ where $u=g(x)$
But I've noticed another way that it is done, where $g'(x)$ is unnecessary.  It seems to involve looking at the problem from another perspective.  Is this correct?
$\int (2x+3)^3xdx = \int (2x+3)^3 \cdot \frac{(2x+3)-3}{2}dx$
I can now integrate with respect to $(2x+3)$ instead of $x$.  If $u=(2x+3)$, then
$$\begin{align*}
\int (2x+3)^3 \cdot \frac{(2x+3)-3}{2}dx&=\int (u)^3\frac{u-3}{2}du\\
&=\int\frac{u^4}{2}-\frac{3u^3}{2}du\\
&=\frac{1}{10}u^5-\frac{3}{8}u^4+c\\
&=\frac{1}{10}(2x+3)^5-\frac{3}{8}(2x+3)^4+c
\end{align*}$$
Is that correct?
I see that the solution would be to see that
$g(x)=u=2x+3$
$g'(x)=\frac{du}{dx}=2$
$dx=\frac{du}{2}$
I can follow this.  However, the last of the three lines there bothers me.  I would like to avoid manipulating derivatives like quotients.  It's something that has caused me problems before.  I avoid using something when I don't understand how it really works.  Does that make sense?
EDIT:
I'm thinking that
$\int f(g(x))dx=\lim_{n \to \infty} \sum_{i=1}^n f((g^{-1}(x))_i^*) \cdot \frac{g^{-1}(b)-g^{-1}(a)}{n}$
..I'm thinking out loud so that might not be right, but that's the kind of approach I want to take to solve this.
 A: You replaced $dx$ with $du$, but $dx\ne du$, given that $u=2x+3$. 
EDIT: Going up to your first paragraph, the $g'(x)$ is important, because it reminds you that you are really doing a chain rule. $(f(g(x)))'=f'(g(x))$ isn't correct; you have to have $(f(g(x)))'=f'(g(x))g'(x)$. And it's exactly that $g'(x)$ that you miss when you replace $dx$ with $du$. Don't think of it as treating a derivative like a fraction; think of it as a bookkeeping device to help you keep track of the $g'(x)$ that has to be there because of the chain rule. 
A: You can always check an answer to an integration problem by differentiating. If you differentiate your result, you will get $(2x+3)^4-3(2x+3)^3$. (If you get something else, you forgot to use the Chain Rule.)
Take out the common factor of $(2x+3)^3$. 
We get $(2x+3)^3((2x+3)-3)$, which is $(2x+3)^3(2x)$, not what you were trying to integrate.
Essentially, you were doing a substitution: in fact you wrote $u=2x+3$ explicitly.
You just didn't bother with the $du=2\,dx$, or equivalently $dx=\frac{du}{2}$. That accounts for the fact that your answer is off by a factor of $2$.
Remark: Do remember that one can always check by differentiating! When one does a complicated integration, slippage (a missing minus sign, a wrong constant) is all too frequent, at least for me. Differentiating will almost always detect an error.
