This is very basic, but you have to start somewhere I guess:
When solving the following first order linear equation: $$ \int \frac{\mathrm{d}x}{\mathrm{d}t}*\frac{1}{x} dt = \int k \mathrm{d}t $$
$k$ is not dependent on $t$, but a constant (I guess)
I don't get where the $\frac{\mathrm{d}x}{\mathrm{d}t}$ part goes. In the solution in my book it states it is equal to
$\ln(x(t)) = kt+c_0$ with $c_0$ an integration constant
But it seems to me integrating $\frac{\mathrm{d}x}{\mathrm{d}t}$ should result in an extra $x$ somewhere
This is the "physicist way", simplify with $\mathrm{d}t$:
$$\frac{\mathrm{d}x}{\mathrm{d}t} \frac{1}{x} \mathrm{d}t=\frac{1}{x} \frac{\mathrm{d}x}{\mathrm{d}t} \mathrm{d}t=\frac{1}{x}\mathrm{d}x\frac{\mathrm{d}t}{\mathrm{d}t}=\frac{1}{x}\mathrm{d}x$$
This is not correct mathematically, but it's really spectacular.
But you can use integrating by substitution, to be mathematically correct:
$$\int\frac{\mathrm{d}x}{\mathrm{d}t} \frac{1}{x} \mathrm{d}t$$ Substitute $u=x(t)$, so $\mathrm{d}u=\frac{\mathrm{d}x}{\mathrm{d}t}\mathrm{d}t$: $$\int\frac{1}{u}\mathrm{d}u=\int \frac{1}{x}\mathrm{d}x$$