Solve: $$\int x(2-3x)^{11} \, dx $$
The book I am following uses a weird technique to solve this. I am having trouble understanding why it works.
Let: $$ u = 2-3x$$ $$ x = \frac{2-u}{3} $$ $$ \frac{du}{dx} = -3 $$
Book Technique: $$dx = - \frac{du}{3}$$
$$ \int \frac{2-u}{3} u^{11} - \frac{du}{3}$$ $$ -\frac{1}{9} \int 2u^{11} - u^{12} du$$
My problem with this is that dy/dx is not a fraction but the limit of one as such the terms dy and dx on there own are in a sense meaningless and cannot be manipulated algebraically. For this reason, I solved it with a more formal technique:
$$ \frac{dy}{du} = (\frac{2-u}{3})(u^{11}) = \frac{2u^{11}}{3} - \frac{u^{12}}{3} $$
Given, the reverse chain rule (i.e u-substitution)
$$ u = g(x), y = f(u) $$ $$ \int \frac{du}{dx} \frac{dy}{du} dx = \int \frac{dy}{du} du$$
so:
$$ -\frac{1}{3} \int -3 x(2-3x)^{11} dx = \int \frac{dy}{dx} du $$ $$ -\frac{1}{9} \int 2u^{11} - u^{12} du $$
My question is thus:
The technique used by the book is at most an approximation, i.e by assuming that $dy/dx$ is fraction we are taking an assumption that will not always hold, but in this case it seemingly does. Is this correct way of looking at why the 'book technique' works?