proof that $\frac{dx}{dt}=f(x)\implies\int \frac{1}{f(x)}dx=\int dt$ I have for a long time been using the rule that $$\frac{dx}{dt}=f(x)\implies\int \frac{1}{f(x)}dx=\int dt$$
I have always "derived" it by multiplying the infinitessimals in the typical way, and then adding $\int$.
However, the infinitessimals are not separately defined in standard calculus, so that derivation makes no sense. 
What is an actual good derivation/proof, that this rule holds? (preferably using theorems, without using $\epsilon$'s and $\delta$'s, unless necessary).
 A: $$\begin{array}{rcll}
\dfrac{\mathrm dx}{\mathrm dt} &=& f(x) & \text{assumption} \\
\dfrac{\mathrm dt}{\mathrm dx} &=& \dfrac1{f(x)} & \text{chain rule} \\
t &=& \displaystyle \int \dfrac1{f(x)} \ \mathrm dx & \text{definition of primitive} \\
\displaystyle \int \ \mathrm dt &=& \displaystyle \int \dfrac1{f(x)} \ \mathrm dx &\begin{array}{l}\text{derivative of identity is 1}\\\text{definition of primitive}\end{array} \\
\end{array}$$
A: By the chain rule
$$\int \frac{1}{f(x)}dx=\int \frac{dt}{dx}dx=\int dt$$
A: You're implicitly using two functions: one of them is $f$ and the other (here denoted by $h$) satisfies 
$$ h'(t) = f(h(t)). $$
If $f \neq 0$ at some point, we can choose $F$ to be a primitive function of $1/f$. Then 
$$ \left( F(h(t)) \right)' = F'(h(t)) \cdot h'(t) = \frac{1}{f(h(t))} \cdot f(h(t)) = 1, $$
hence 
$$ F(h(t)) = t + C, $$
as needed. In the original notation, this would be $F(x) = t+C$. 

It is easily seen that the same works for ODEs of the form 
$$ \frac{dx}{dt} = \frac{f(x)}{g(t)}. $$
Again, let $h$ be a solution and $F$ be a primitive function of $1/f$. Then 
$$ \left( F(h(t)) \right)' = F'(h(t)) \cdot h'(t) = \frac{1}{f(h(t))} \cdot \frac{f(h(t))}{g(t)} = \frac{1}{g(t)}. $$
Integrating this, we obtain an equation for $h$ equivalent to 
$$ \int \frac{dx}{f(x)} = \int \frac{dt}{g(t)}. $$
