Is it true that $\int\frac{1}{x}\,dx=\ln x\implies\ln x = \frac{x^0}{0}$? I have recently learned that $\int\frac{1}{x}\,dx=\ln x$.
However, by the power law of integration, the integral of $1/x$ is equal to $x^0 / 0$ which is undefined. Therefore, is $x^0 / 0 = \ln x$?
Am I correct in this assumption?
 A: Your observation can be properly understood using a limit. Specifically, $f(x,p)=\int_1^x y^{p} dy$ is a continuous function of $p$, even at $p=-1$. This means that
$$\lim_{p \to -1} \frac{x^{p+1}-1}{p+1} = \ln(x).$$
You could prove this directly with L'Hospital's rule, provided you already knew that $\frac{d}{dx} a^x = a^x \ln(a)$. 
A: The power rule cannot be used when integrating $\int \frac 1x\,dx = \int x^{-1}\,dx$.
I assume you are referring to the fact that for $n\neq -1$, $$\int x^{n} \,dx = \dfrac{x^{n+1}}{n+1}+ C$$  Note that this holds for all integers $n \neq -1$.
So $\dfrac {x^0}0 \neq \ln |x|$, and as you know, $\dfrac {x^0}0$ is not defined.
A: We can define $\log{x}$ explicitly as
$$ \int_1^x \frac{dt}{t}; $$
but why this definition?
Note that, for general powers $x^{n-1}$, we have
$$ \int_1^x t^{n-1} \, dt = \frac{x^n-1}{n}, $$
so we can't just set $n=0$ here to have a sensible answer without getting $0/0$. However, this suggests that
$$ \lim_{n \to 0} \frac{x^n-1}{n} = \log{x}. \tag{1} $$
This is in fact true, even if we define the logarithm as the inverse to the exponential: recall the definition of $e^x$ as a limit,
$$ y = e^x = \lim_{m \to \infty} \left( 1+\frac{x}{m}\right)^m $$
say, then we can do some algebra to find
$$ x= \lim_{m \to \infty} m(y^{1/m}-1), $$
(or just substitute this in and see what happens) and if you set $m=1/n$ you recover the limit (1).
A: A slight, but important correction to make:
$\int\frac{1}{x}\,dx=\ln |x|$ + c
We use the absolute value of x in order to have the same domain for both the right and left hand side.
If we only take the natural log of x, our x values are restricted to being > 0.
Whereas in  $x^-1$, Our x values are only restricted to x !=0
A: all the answers so far use integral representation of $\ln x.$ i will use the fact that $$ \text{ the derivative of } a^x \text{ at } x = 0 \text{ is }\ln a.$$  written in the form $$\lim_{x \to 0} \frac{a^x - 1}{x} = \ln a. $$  switching  the roles of $x$ and $a$ we have $$\lim_{a \to 0} \frac{x^a - 1} a = \ln x. $$ in a less than rigorous way, we have  $$\frac{x^0 - 1} 0 = \ln x  $$
