Why $\int \frac{1}{x}dx = \ln{x}$? I used the search, but for some reason I can not find such a question. This is probably very simple, but not for me...
I want to understand why $\int \frac{1}{x}dx = \ln{x} +\rm  const$?
I'm trying to solve this as $$\int \frac{1}{x}dx = \int x^{-1} dx = \int x^{-1} dx = \frac{x^{-1+1}}{-1+1} + {\rm const}=\frac{x^0}{0} + \rm const=\frac{1}{0} + const$$
How do I uncover uncertainty?
I don't know how to apply knowledge about the logarithm ($x=\log_{a} b \Leftrightarrow a^x=b$) in this case.
 A: It may help to think about definite integrals viz.$$\epsilon\ne0,\,x>0\implies\int_1^xt^{\epsilon-1}dt=\frac{x^\epsilon-1}{\epsilon},$$so$$x>0\implies\int_1^xt^{-1}dt=\lim_{\epsilon\to0}\frac{x^\epsilon-1}{\epsilon}=\ln x\cdot\lim_{\epsilon\to0}\frac{e^{\epsilon\ln x}-1}{\epsilon\ln x}.$$But this last limit is the definition of the $y$-derivative of $e^y$ at $0$, since $h:=\epsilon\ln x\to0$ as $\epsilon\to0$. This limit is $1$, completing the proof.
A: So the power rule breaks down because of a division by zero, however there is still continuity as you move the exponent through $-1$, provided that you are considering a definite integral, or at least an integral with one limit of integration pinned down. Conventionally we do this by pinning the lower limit to $1$ and allowing the upper limit to vary. Thus you look at $\int_1^x y^p dy = \frac{x^{p+1}-1}{p+1}$, with $x>0$.
Now you send $p \to -1$, which amounts to computing $\frac{d}{dp} x^{p+1}$ at $p=-1$. Now the connection to the exponential comes from the need to differentiate an exponential expression with respect to its exponent.
There is no continuity in the usual statement of the power rule for indefinite integrals, i.e. $\int x^p dx = \frac{x^{p+1}}{p+1}$, because this corresponds to $\int_0^x y^p dy$ which is not a finite integral when $p \leq -1$.
A: for $x>0$ we have ...
$$ e^{\ln x} = x \implies \frac{d}{dx}  e^{\ln x} = 1 $$
Calculate the same derivative using chain rule ...
$$ \frac{d}{dx}  e^{\ln x} = e^{\ln x}  \bigg (  \frac{d}{dx}  \ln x \bigg )
\\ = x \bigg ( \frac{d}{dx}  \ln x \bigg ) $$
Equating the two expressions you get
$$   \frac{d}{dx}  \ln x  =\frac 1x$$
The result now follows from the fundamental theorem of Calculus.
