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.