First I apologise because my English is broken. But... If the following picture's integral :
is $\int f(x) dx$, how can we write this integral Integral2
For example , I needed did this :
$\int x^{-1} dx = Ln(x)$
$\int x^{-1} = \frac{Ln(x)}{dx}$
$\int x^{-1} \Delta x = \frac{Ln(x)}{dx}\Delta x$
But I think this is not allowed because dx is in the limit (to zero).
Suppose you are integrating over an interval $[a, b]$ and your rectangles in the second figure have a width $\Delta x$. For simplicity, let us assume $n = \frac{b - a}{\Delta x}$ is an integer. Firstly, note that you have \begin{align*} \int_{a}^{b} \frac{1}{x} \ \mathrm{d}x = \sum_{i = 1}^n \int_{a + (i-1)\Delta x}^{a + i \Delta x} \frac{1}{x} \ \mathrm{d}x. \end{align*} This is simply rewriting the overall area as the sum of areas under $[a, a + \Delta x], [a + \Delta x, a + 2\Delta x], \dots [a + (n-1)\Delta x, a + n\Delta x]$ where $a + n\Delta x = b$, by definition.
If you want to further simplify the expression, you can use several approaches. You can upper or lower bound the summation as follows. Note that $1/x$ is a strictly decreasing function which means that in an interval $[a + (i-1)\Delta x, a + i\Delta x]$ for any $i \in \{1,2,\dots, n\}$, $\dfrac{1}{a + i\Delta x} \leq \dfrac{1}{x} \leq \dfrac{1}{a + (i-1)\Delta x}$. Using this, we can obtain an upper bound as follows: \begin{align*} \int_{a}^{b} \frac{1}{x} \ \mathrm{d}x & = \sum_{i = 1}^n \int_{a + (i-1)\Delta x}^{a + i \Delta x} \frac{1}{x} \ \mathrm{d}x \\ \leq & \sum_{i = 1}^n \int_{a + (i-1)\Delta x}^{a + i \Delta x} \frac{1}{a + (i-1)\Delta x} \ \mathrm{d}x \\ \leq & \sum_{i = 1}^n \frac{1}{a + (i-1)\Delta x} \int_{a + (i-1)\Delta x}^{a + i \Delta x} \ \mathrm{d}x \\ \leq & \sum_{i = 1}^n \frac{\Delta x}{a + (i-1)\Delta x}. \end{align*}
You can similarly obtain a lower bound of the form $\displaystyle \sum_{i = 1}^n \frac{\Delta x}{a + i\Delta x}$. You can also use the mean value theorem to obtain an exact expression. Using the mean value theorem, you can conclude that for all $i \in \{1,2,\dots, n\}$, there exists $\alpha_i \in [a + (i-1)\Delta x, a + i\Delta x]$ such that $\displaystyle \int_{a + (i-1)\Delta x}^{a + i \Delta x} \frac{1}{x} \ \mathrm{d}x = \frac{\Delta x}{\alpha_i}$. Consequently, you can write \begin{align*} \int_{a}^{b} \frac{1}{x} \ \mathrm{d}x = \sum_{i = 1}^n \frac{\Delta x}{\alpha_i}. \end{align*} However, this might not be the most useful approach since it is difficult to compute $\alpha_i$'s in closed form.