This is essentially a question about an answer I received at - Integrating a function using u-substitution. Initially, I thought I understood it, but looking at it, I do have uncertainties.
I just want to understand if I got the right idea solving, so I will be listing the steps and my uncertainties alongside with them:
Quoting steps from amWhy:
$$\int\ \frac {e^{2x}-6e^x}{e^x+2}\ dx$$
Let $\displaystyle e^x + 2 = u$.$^{\color{blue}{(1)}}$
So $\displaystyle e^x = u-2$, and so $\displaystyle e^{2x} = (u-2)^2$.
$\color{blue}{(1)}$ Now $\displaystyle e^x \ dx = du \iff \ dx = \frac{du}{e^x}$, but recall from above, that $\displaystyle e^x = u-2$.
So in fact, $\displaystyle \color{blue}{dx = \frac{du}{u-2}}$.
That gives us the integral
$$\begin{align} \int \frac{(u-2)^2 - 6(u-2)}{u\cdot \color{blue}{(u-2)}}\ >\color{blue}{du} & = \int \frac{(u-2) - 6}{u}\,du \\ & = \int \frac{u-8}{u}\,du >\\ &= \int \left(1- \frac 8u\right)\ du = \end{align}$$
I understand the part where we take $dx=\frac{du}{e^x}$ turn it into $dx=\frac{du}{u-2}$, what I don't understand is why we ACTUALLY MULTIPLY the whole equation of $\int \frac{(u-2)^2 - 6(u-2)}{(u)}$ by $\frac{du}{u-2}$.
I will give you an example of why it doesn't make sense to me:
Say we have:
$\int \frac{(5x^4)}{x^5+5}dx$ >> $u=x^5+5$ and THIS: $du=5x^4 dx$ or just $\frac {du}{5x^4}=dx$ >> $\int \frac{1}{u}du$
Now we don't actually multiply the whole equation by $\frac {du}{5x^4}$, it just turns into $du$, so why did we do it in the first example?
Thanks in advance for any clarification!
Let us suppose we wished to integrate $$\int{\frac{5x^4}{x^5+5}dx}$$
You suppose we take $u=x^5+5$, then $\frac{du}{dx}=5x^4\to du=5x^4 dx$
In this particular case, it makes sense to leave the $5x^4$ as-is, as it is involved in the integral anyway So $$\int{\frac{{5x^4}}{x^5+5}dx}$$ becomes $$\int{\frac 1u du}=\ln|u|+C$$ Because $du=5x^4dx$ and $u=x^5+5$
You may have noticed that $5x^4$ is the derivative of $x^5+5$, as a shortcut you can say: $$\int{\frac{f'(x)}{f(x)}dx}=\ln|f(x)|+C$$
Let me now show you another integral where $u$-substitution is used:
Suppose we want: $$\int{\frac{1}{x^2+1}dx}$$
Let $x=\tan(u)$, then $\frac{dx}{du}=\sec^2(u)\to dx=\sec^2(u)du$
Now since we have $dx$ in terms of $du$, we can replace it so: $$\int{\frac{1}{x^2+1}\sec^2(u)du}$$ And we can replace the $x$ with $\tan(u)$, so: $$\int{\frac{1}{\tan^2(u)+1}\sec^2(u)du}$$ And since $\tan^2(u)+1=\sec^2(u)$, we get $$\int{\frac{sec^2(u)}{sec^2(u)}du}=\int{1 du}=u+C$$
In the first example you gave, this process is the same, you simply replace $dx$ by the equivalent $\frac{1}{u-2}du$