How can I prove the integral $ \int_{1}^{x} \frac{1}{t} \, dt $ is $\ln x $ with this approach? I have been trying to find a proof for the integral of $ \int_1^x \dfrac{1}{t} \,dt $ being equal to $ \ln \left|x \right| $ from an approach similar to that of the squeeze theorem.
Is it possible to calculate the area under the curve $ f(x) = \dfrac{1}{x} $ as in the picture shown below? You may notice that both sums of the areas should converge to $\ln(x)$ as the base of the rectangles gets smaller and smaller. We approach from above and from below to get a limiting argument of the form:
Area from below the curve as in the second graph $ \leq \int_a^b \dfrac{1}{x} \,dx \leq$ Area from above the curve as in the first graph
The limiting argument would be to keep calculating and adding the areas of the rectangles which bases get smaller and thus showing that this amount is $\ln(x)$.

How could I proceed in this way?
 A: The integral you present is often taken to be the definition of the $\log$-function: 
$$L(x) \equiv \int_1^x \frac{dt}{t}$$
If we use this as the definition then the challenge becomes to show that this is the good old $\log$ we know and love, i.e. that it satisfies the usual $\log$ properties and that it is the inverse function of the exponential function $e^x$ where $e$ is Euler's constant.
By a change of variables $z = ty$ to the integral above we can derive the $\log$-property $L(x) + L(y) = L(xy)$ as follows
$$L(x) = \int_y^{xy}\frac{dz}{z} = \int_1^{xy}\frac{dz}{z} - \int_1^y \frac{dz}{z} = L(xy) - L(y)$$
Likewise a change of variables $z=t^\alpha$ to the integral $L(x^\alpha)$ gives us $L(x^\alpha) = \alpha L(x)$ so $L$ satisfies both of the well known $\log$-properties. Since $L(x)$ is strictly increasing on $(0,\infty)$ it has an inverse function $E(x)$ which satisfies
$$L(E(x)) + L(E(y)) = L(E(x)E(y)) \implies E(x+y) = E(x) E(y)$$
where we used the property $L(E(x)) = x$. This functional equation has the (continuous) solution $E(x) = {\bf e}^x$ where ${\bf e}\equiv E(1)$ is a constant that must satisfy
$$L({\bf e}) = 1\implies \int_1^{{\bf e}}\frac{dt}{t} = 1$$
The final thing to show is that this ${\bf e}$ is indeed Euler's constant $e$. This can be done by using the definition $e \equiv \lim\limits_{n\to\infty}\left(1+\frac{1}{n}\right)^{n}$ to get
$$L(e) \equiv \lim_{n\to\infty}nL\left(1+\frac{1}{n}\right) = \lim\limits_{n\to\infty}n\int_1^{1+\frac{1}{n}}\frac{dt}{t} = 1$$
and since $L$ is injective $e={\bf e}$.
A: As has been pointed out above, there are many different ways of defining $\log x$. I'm going to assume here that you're taking it to be the inverse of $\exp x$, which is itself defined (again, there are many possible equivalent ways of defining it) as $y(x)$ for the unique solution of the differential equation $y' = y$ with $y(0) = 1$. It's straightforward to prove directly from that definition that $\exp$ is a increasing function on the real line and nonzero everywhere. (If you're not, the trick is to prove the addition formula $\exp(a + b) = \exp a\exp b$ by considering the function $y(x) = \exp(x + a)/\exp a$ and invoking uniqueness.)
Given that, the function $y(x) = \exp x$ has $\frac{y'}{y} = 1$. Integrating then gives 
$$x = \int_0^x \frac{y'}{y} \, dt= \int_1^{y(x)} \frac{dt}{t}.$$
Replacing $x$ with $y^{-1}(x)$ then gives
$$y^{-1}(x) = \int_1^x \frac{dt}{t}.$$
A: Assume you're starting with definition of $\ln$ as inverse of $\exp$ (and that you already know $\exp$ is differentiable and $\exp' = \exp$)...
By Inverse Function Theorem, $\ln$ is differentiable. From $\exp(\ln x) = x$ and Chain Rule, $\ln'(x) = 1/x$. By Fundamental Theorem of Calculus, $\int_1^x (1/t) dt = \ln x - \ln 1 = \ln x$.
A: make the substitution $\frac{1}{x}=e^u$, and see that the integral $\int\frac{1}{x}dx = -u$
A: I don't know if this helps but following your suggestions I found a way to approach $\ln(1+x)$ using rectangles that have smaller and smaller bases. I don't think this constitutes a proof but shows that adding these rectangles equals $\ln(1+x)$. The result is a function that is nothing more but an infinite sum of infinitesimally small rectangles that mimics perfectly $\ln(1+x)$ for values where $x>-1$. The approach is as follows:
Start by taking the area of a rectangle that has length $x$ and height $f(x)$. The area is $f(x).x$. This is too large and a bad approximation. So break the interval from $1$ to $1+x$ by two intervals of length $\dfrac{x}{2}$. Then the area would be, starting from the left: 
$f(1). \dfrac{x}{2} + f(1+\dfrac{x}{2}). \dfrac{x}{2} = \dfrac{x}{2} + \frac{2}{1+x}.\frac{x}{2} = x(\dfrac{1}{2}+\dfrac{1}{2+x}) $
Then, in a third iteration, breaking $x$ into three intervals and calculating the area of the three rectangles, we would get:
$f(1). \dfrac{x}{3} + f(1+\frac{x}{3}). \dfrac{x}{3} + f(1+2\frac{x}{3}). \dfrac{x}{3} = \dfrac{x}{3} + \frac{3}{1+x}.\frac{x}{3} + \frac{3}{1+2x}.\frac{x}{3}    =  x(\dfrac{1}{3} + \dfrac{1}{3+x} + \dfrac{1}{3+2x}) $
As you can see, a pattern starts to emerge. As we break the $x$ interval into smaller and smaller parts and sum the area of the smaller rectangles we approximate better the area under $f(x)=\frac{1}{x}$. This would give a formula as follows:
$\sum_{k=0}^{n-1}(\frac{x}{n+xk})$
where $k$ represents the begining points of the bases of the rectangles and $n$ represents by how many chunks we break up $x$. Now, take the limit of $n$ going to infinity to get:
$$\lim_{n\to\infty} \sum_{k=0}^{n-1}(\frac{x}{n+xk})$$
and you will notice that this expression is equivalent to $\ln(1+x)$. I don't know how to prove this mathematically, but if you plot both expressions on a graph you will notice that even with a small value of $n$ we approach quite well the $\ln(1+x)$ function. As $n$ increases both functions seem to converge everywhere.
I don't know how the infinite sum relates to the Taylor approximation of $\ln(1+x)$. I suspect that equating this expression or using other tricks might show that both expressions are equivalent, but I have not succeeded at this. Anybody out there has any suggestions?
