Can someone explain why this is true? $\int f(x) g'(x)\,dx = \int f(x)\,dg$ I'm a software engineer (have been for 20+ years) and have over the past few years taken quite an interest in math. I would appreciate it if anyone is able to help me out with my question.

I understand the following:

$$\int g’(x)\,dx = \int dg$$

This makes sense to me.

$$\int g’(x)\,dx = g(x)+c$$
$$\int dg = g(x) + c$$

That being the case, $\displaystyle \int g'(x)\,dx = \int dg$.

As far as I know, this is also true:

$$\int f(x)g’(x)\,dx = \int f(x)\,dg$$

What I don't completely understand is why you can put $f(x)$ in each of these integrals and they continue to be equal. I mean, I understand that if $\int g’(x)\,dx = \int dg$, then I should be able to swap $g’(x)\,dx$ for $dg$ (or vice versa), but I'm trying to visualize how $f(x)$ doesn't somehow throw this off.

Just reading it, I am looking for the integral of $f(x)$ multiplied by $g’(x)$ with respect to $x$ and the integral of $f(x)$ with respect to $g$.

Can someone explain the logic of this to me and/or point me to an applicable proof? I've Googled around to see if I can find a good explanation, but I must not be describing my question in the search query well enough (hopefully I am doing an OK job describing it here).
What you should read is that, by definition, $dg=g'(x)\,dx$. This relation holds independently of integration.

For example, you can use it in the chain rule,

$$\frac{dg}{dt}=g'(x)\frac{dx}{dt}.$$

[If you want to be picky, $\dfrac{dg(x(t))}{dt}=g'(x(t))\dfrac{dx(t)}{dt}$.]
As Yves Daoust writes, this is from the simple fact:

$$\frac{dg}{dt} \cdot dt = /\text{ cancel dt }/=dg$$

This is independent of integration.