What does $dx$ mean in differential form? This question relates to this post. 
From what I know in calculus and standard analysis, strictly speaking, there is no meaning of $dx$. It only makes sense when combining with another $d$, e.g. $df/dx$ as derivative or integration, e.g. $\int f(x) dx$. The $dx$ in derivative or integration has a definite meaning inside the definition of derivative or integration, respectively. 
However, in differential form, it is given as
$\omega = \frac{ \omega_{\mu_1,\cdots, \mu_r}}{r!} dx^{\mu_1} \wedge\cdots\wedge dx^{\mu_r} $
where $dx$ appears explicitly. What does $dx$ mean in differential form? Physicist usually say it is infinitesimal. However, infinitesimal does not mean anything in standard analysis, or I am completely mistaken?
 A: $dx_1$ is a differential 1-form (aka a covector field) which associates to each point in space a linear map from $\mathbb{R}^n \to \mathbb{R}$.  The action of this linear map is to take a vector and spit out its component in the $x_1$ direction.  In other words, it is the covector $\begin{bmatrix} 1&0&0&...&0 \end{bmatrix}$.
The covector fields $dx_i$ span all covector fields in the algebra of these guys over real valued functions, just because you can write
$$\begin{bmatrix} f_1(x)&f_2(x)&f_3(x)&...&f_n(x) \end{bmatrix}$$ as $$f_1(x)dx_1+f_2(x)dx_2 + ...+ f_n(x)dx_n$$
The way you integrate a differential one form $\omega$ along a curve is pretty simple.  Given a curve $\gamma: [0,1] \to \mathbb{R}^n$, partition it into $k$ pieces.  Then you can form the sum 
$$\sum_{i=1}^k \omega\big|_{\gamma(\frac{i}{k})}(\gamma(\frac{i}{k}) - \gamma(\frac{i-1}{k}))$$
The limit of this sum as $k \to \infty$ is the integral of the one form.  Alternatively you could have plugged in actual tangent vectors at each point, instead of the approximate tangent vectors I used, but I think it is somewhat easier to conceptualize what is going on this way.
Why are these reasonable things to look at?  Why would anyone ever think of integrating such a thing?  Answer:  because the derivative of a function $f:\mathbb{R}^n \to \mathbb{R}$ IS a covector field, and we certainly want to be able to integrate a derivative over a curve, and have a fundamental theorem of calculus.  In fact, $dx_1$ is just the derivative of the coordinate function $f(x_1,x_2,...,x_n) = x_1$. If you think about how you would sensibly integrate the derivative of a function, you will probably recover my notion of integration above.
A: My first answer just answered the question directly, but I would like to take a little bit of time to explore this circle of ideas.
Let $f(x,y) = x^2y$.
The derivative of a function gives the best linear approximation to the function at a point.  This remains true in higher dimensions.  The derivative of the function $f$ above is 
$df = \begin{bmatrix} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y}\end{bmatrix} = \begin{bmatrix} 2xy & x^2\end{bmatrix}$.
The conceptual meaning of the derivative is this:  
$$f(x+\Delta x, y+\Delta y) \approx f(x,y) + df(\begin{bmatrix} \Delta x \\ \Delta y\end{bmatrix}) = x^2y + \begin{bmatrix} 2xy & x^2\end{bmatrix}\begin{bmatrix} \Delta x \\ \Delta y\end{bmatrix}  = x^2y+2xy\Delta x + x^2\Delta y$$
In other words, at each point $(x,y)$ the derivative is a linear map which takes a small change $\begin{bmatrix} \Delta x \\ \Delta y\end{bmatrix}$ away from the point $(x,y)$ and returns the approximate change in $f$ resulting from that.
Now say someone told me that a certain function $g$ with $g(0,0)=0$ had derivative $\begin{bmatrix} y\cos(xy) & x\cos(xy)\end{bmatrix}$, and I wanted to figure out what $g$ was.  In this case I could probably just solve the differential equations $\frac{\partial g}{\partial x} = y\cos(xy)$ and $\frac{\partial g}{\partial y} = y\cos(xy)$ by inspection, but this would not always be possible.
Let us stick to the somewhat easier problem of approximating $g(1,1)$.  Here is my idea for doing that:  I will pick a path from $(0,0)$ (whose value I know) to $(1,1)$.  I will split that path up into millions of vector changes.  Then I will use what I know about the derivative to approximate the change in $g$ over each of those small changes and add them up.  This should give me a pretty reasonable approximation.
In this case, I can see that I can pick the path $\gamma:[0,1] \to \mathbb{R}^2$ given by $\gamma(t) = (t,t)$.  Splitting this into $k$ pieces, I have the following approximations:
$$g(\frac{1}{k},\frac{1}{k}) \approx g(0,0) + dg|_{(0,0)}\left(\begin{bmatrix} \frac{1}{k} \\ \frac{1}{k}\end{bmatrix}\right)$$. 
So then 
$$g(\frac{2}{k},\frac{2}{k}) \approx g(\frac{1}{k},\frac{1}{k}) + dg|_{(\frac{1}{k},\frac{1}{k})}\left(\begin{bmatrix} \frac{1}{k} \\ \frac{1}{k}\end{bmatrix}\right) \approx g(0,0) + dg|_{(0,0)}\left(\begin{bmatrix} \frac{1}{k} \\ \frac{1}{k}\end{bmatrix}\right) + dg|_{(\frac{1}{k},\frac{1}{k})}\left(\begin{bmatrix} \frac{1}{k} \\ \frac{1}{k}\end{bmatrix}\right)$$.
Continuing on in this way , we will see that 
$$g(1,1) \approx g(0,0) + \sum_{i=0}^k dg\big|_{\frac{i}{k}}\left(\begin{bmatrix} \frac{1}{k} \\ \frac{1}{k}\end{bmatrix}\right)$$
It makes sense to give some name to this process.  We define the limit of the sum above to be the integral of the covector field $dg$ along the path $\gamma$.  Refer to my other post for the general definition, instead of just a particular example like this.
So far we have defined the integral only for derivatives of functions, and we have defined it exactly in such a way that the following fundamental theorem of calculus holds:
$$g(P_1) - g(P_0) = \int_\gamma dg$$ for any path $\gamma$ from $P_0$ to $P_1$.  But the definition of the integral never used the fact that we were integrating the derivative of a function:  it only mattered that we are integrating a covector field (i.e. a gadget which eats change vectors and spits out numbers).  So we can use exactly the same definition to give the integral of a general covector field $\begin{bmatrix} f(x,y) & g(x,y)\end{bmatrix}$, which may or may not be the differential of a function.  
(There are certainly covector fields which are not derivatives of functions.  For example, $\begin{bmatrix} x & x\end{bmatrix}$ could not be the differential of a function, for if it were we would have $\frac{\partial f}{\partial x} = x$ and $\frac{\partial f}{\partial y} = x$.  But then the mixed partials of $f$ would not be equal, contradicting Clairout's theorem.)
$dx$ is the constant covector field $\begin{bmatrix} 1 & 0\end{bmatrix}$, and $dy$ is the constant covector field $\begin{bmatrix} 0 & 1\end{bmatrix}$.  So we can write any covector field $\begin{bmatrix} f(x,y) & g(x,y)\end{bmatrix}$ as $f(x,y)dx + g(x,y)dy$.  Integrating this thing along a curve is PRECISELY what you defined as a line integral in your first multivariable calculus course.  
