Explain the integral of $1/x = \ln |x| + \mathrm{C}$ graphically as sum of area? I am unable to interpret the integral
$$\int {1\over x}{\rm d}x=\ln|x|+\mathrm{C}$$
Graphically as area under the curve of $1/x$ (as the definition of the integral). Can somebody please illustrate/explain.
I understand this fact that derivative of $\ln(x)$ is $1/x$ and we take absolute value to make the support extend to entire real line, $x \neq 0$.
 A: Another reason is
that,
 if we start with
the functional equation
for the logarithm,
$f(xy)
=f(x)+f(y)
$,
we end up with the
integral of
$\frac1{x}$.
Here's how it goes:
We have
$f(xy)
=f(x)+f(y)
$,
and we want to find
$\lim_{h \to 0} 
\dfrac{f(x+h)-f(x)}{h}
$.
In what follows,
I assume that
$f(x)$ is as smooth as needed.
In particular,
I assume that
$f'(x)$
exists for
$x > 0$.
First,
putting $x=y=1$,
we get
$f(1) = f(1)+f(1)$,
so that
$f(1) = 0$.
Then,
$f(x+h)
=f(x(1+h/x))
=f(x)+f(1+h/x)
$,
so that
$f(x+h)-f(x)
=f(1+h/x)
$.
Therefore
$\begin{array}\\
\dfrac{f(x+h)-f(x)}{h}
&=\dfrac{f(1+h/x)}{h}\\
&=\dfrac1{x}\dfrac{f(1+h/x)}{h/x}\\
&=\dfrac1{x}\dfrac{f(1+h/x)-f(1)}{h/x}
\quad\text{(since }f(1) = 0)\\
\end{array}
$.
Letting
$h \to 0$
on both sides,
we get
$f'(x)
=\dfrac{f'(1)}{x}
$,
or
$\begin{array}\\
f(x)
&=f(x)-f(1)\\
&=\int_1^x f'(t)dt\\
&=\int_1^x \dfrac{f'(1)dt}{t}\\
&=f'(1)\int_1^x \dfrac{dt}{t}\\
\end{array}
$
If we choose
$f'(1)
= 1
$,
which is a natural choice,
we get
$f(x)
= \int_1^x \dfrac{dt}{t}
$,
which is one of the reasons
this is called the
natural logarithm.
By choosing different values
for $f'(1)$,
we can get the other logs:
base 10, base 2,
and so on.
As if often the case
in my answers,
nothing original,
and I've said this before,
but if the question isn't
being flagged as a duplicate,
then my answer shouldn't either.
A: Let's define a function $f$ as $f(x) = \int_1^{x} \frac{1}{t} dt$, $x>0$.
It is easy to see show the following properties:

Property (1)
The function $f$ is increasing.
To show this, assume $x_2>x_1>0$, and form the difference $f(x_2)-f(x_1)=\int_{x_1}^{x_2} \frac{dt}{t}$.  Note that this difference is strictly positive since the integrand is also strictly positive. Thus, $f(x_2)>f(x_1)$ whenever $x_2>x_1$, which implies $f$ is strictly increasing.

Property (2)
The function $f$ is differentiable with $f'(x)=\frac{1}{x}$.
This follows immediately from the fundamental theorem of calculus.  Note that this implies automatically that $f$ is continuous for $x>0$.

Property (3)
The limit $\lim_{x \to \infty} f(x)=+\infty$ 
We can easily verify that $\lim_{x \to \infty} f(x)=+\infty$ since for $n>1$
$$\int_1^{2^n} \frac{1}{t} dt=\int_{1}^{2} \frac{1}{t} dt+\int_{2}^{4} \frac{1}{t} dt+\cdots +\int_{2^{n-1}}^{2^{n}} \frac{1}{t} dt$$ 
$$\ge \int_{1}^{2} \frac{1}{2} dt+\int_{2}^{4} \frac{1}{4} dt+\cdots +\int_{2^{n-1}}^{2^{n}} \frac{1}{2^n} dt=\frac12 (2-1)+\frac14 (4-2) + \cdots + \frac{1}{2^{n}}(2^n-2^{n-1})=\frac{n}{2}$$
which tends to infinity as $n \to \infty$.

Property (4)
The limit $\lim_{x \to 0} f(x)=-\infty$
Examine the integral
$$f(\frac{1}{n})=-\int_{\frac{1}{n}}^1 \frac{dt}{t}$$
and substitute $u=\frac{1}{t}$.  Then, we have
$$-\int_{\frac{1}{n}}^1 \frac{dt}{t}=-\int_1^n \frac{du}{u}$$
But from property (3), this integral tends to $+\infty$ and thus $f(\frac1{n})$ tends to $-\infty$ as $n \to 0$.

Property (5)
The function $f$ satisfies $f(x^n) = nf(x)$.  
We know by definition that
$$f(x^n) = \int_1^{x^n} \frac{1}{t} dt$$
Then, by substituting $u=t^n$, with $du=nt^{n-1}dt$, and the limits going from $1$ to $x$ reveals
$$\log (x^n)  =\int_1^{x} \frac{1}{u^{\frac{1}{n}}} nu^{\frac{1-n}{n}}du=\int_1^{x} n\frac{1}udu=nf (x)$$


NOTE:
The other well-known properties for the log, such as of a product and quotient rules, follow from Property (5).

We can also show that the inverse function, call it  $exp(x)$, for $f(x)$ has all of the properties of an exponential function.  
And finally, if we define the number $e=exp(1)$ such that $f(e)=1$, then we show that
$$e=\lim_{x\to 0}\left(1+x\right)^{1/x}$$
since $f'(1)=1=f(e)$ and also
$$\begin{align}
f'(1)&\equiv\lim_{h\to 0}\frac{f(1+h)-f(1)}{h}\\\\
&=\lim_{h\to 0}f\left((1+h)^{1/h}\right)\\\\
&=f(e)
\end{align}$$
which by continuity of $f$ implies $e=\lim_{h\to 0}\left((1+h)^{1/h}\right)$  and we are done!
A: If we define $F(x)=\int_1^x f(t)dt$ and differentiat it we see that $F'(x)=f(x) $ also $F'(ax)=f(x)$ then differenc of $F(x)$ and $F(ax)$ must be constant. After some calculation we find  that $F$ is a logarithm that called natural logarithm. So $Ln(x)$ is the area of under of curve $y=\frac{1}{x}$ from $1$ to $x$ which is positive for $x>1$ and negative for $x<1$ and obviously $0$ for $x=1$.
A: You ask to demonstrate the result as a sum of areas, so ... Consider approximating the integral with $n$ "left-handed" Riemann rectangles with congruent bases, observing
$$\int_{y}^{xy}\frac{dt}{t} \;\approx\; \sum_{k=0}^{n-1}\frac{\frac{1}{n}(xy-y)}{y+\frac{k}{n}(xy-y)} = \sum_{k=0}^{n-1}\frac{\frac{1}{n}(x-1)}{1+\frac{k}{n}(x-1)} \approx \int_1^{x} \frac{dt}{t}
$$
Of course, as $n\to\infty$, the "$\approx$"s become "$=$"s, so that
$$\int_{y}^{xy}\frac{dt}{t} = \int_1^{x} \frac{dt}{t}$$
Writing $L(x) := \int_1^x dt/t$, this implies
$$L(x) + L(y) = L(xy)$$
so that the functional equation for the logarithm is satisfied by $L$. Determining that $L$ must be, specifically, the natural logarithm is left as an exercise to the reader.
