Just wanted to do some quick sanity checks on a few problems. Also, why does u substitution work on a high level. When I get to a line like $du = -3y^2 \, dy$, what does the $du$ and the $dy$ individually mean?
- $$\int y^2 ( 4 - y^3)^{\frac{2}{3}}\,dy$$
Let's try $u = 4-y^3$ so $ \frac{du}{dy} = -3y^2$ so $du = -3y^2\,dy$ so $\frac{du}{-3} = y^2\,dy$ $$ \frac{-1}{3} \int u^{\frac{2}{3}}\,du$$
$$= \frac{-1}{3} \cdot \frac{3}{5} u^{\frac{5}{3}}$$
$$= \frac{-3}{15} (4 - y^3)^{\frac{5}{3}} + C$$
- $$ \int \frac{e^u}{(1-e^u)^2}$$
let's say $v = 1 - e^u$ so $\frac{dv}{du} = -e^u$ so $dv = -e^u \, du$ so $-dv = e^u\,du$
$$ - \int \frac{1}{v^2} \, dv$$
$$ - \int v^{-2} \, dv$$
$$v^{-1} = \frac{1}{1 - e^u} + C$$
Do these look right? What do the $du$ and the $dx$ mean on its own? You can just look at them as differentials right? $du$ represents the change in u based on the slope at x right whereas u is the actual change in $u$ right?
Your results are correct. And also your interpretation of the differentials is essentially correct.
Note that the meaning of these differentials is the same as in the chain rule, from which derives the technique of integration by substitution:
if $F(g)$ is a function such that $\frac{dF}{dg}=f(g)$ and $g=g(x)$ is a function of another variable $x$, than $$ \frac{d}{dx}F(g(x))=f(g(x))g'(x)dx $$
You can see a graphical interpretation of the integration by substitution at: Is there a way to graphically visually integration by substitution?