How to find the indefinite integral of $\frac{1}{x}$? The indefinite integral of $\frac{1}{x}$ is $\ln x$, but I only know this because I know that the derivative of $\ln x$ is equal to $\frac{1}{x}$. 

Is there a way to find that the integral of $\frac{1}{x}$ is $\ln x$, without using the knowledge of the derivative of $\ln x$?

 A: Sort of. Use the substitution $x=e^u\implies dx=e^u du$ (also implies that $u=\ln(x)$). to get $$\int\frac{1}{x}dx=\int e^{-u}\cdot e^udu=\int 1du=u+C=\boxed{\ln(x)+C}$$
A: $x\mapsto \frac {1}{x} $ is continuous at $(0,+\infty) $, so it has an antiderivative.
the antiderivative which is zero for $x=1$ is called $\ln $.
By definition
$$\ln (x)=\int_1^x\frac {dt}{t} $$
for $x>0$.
A: If you define a function $L(x)$ as the integral
$$L(x)=\int_1^x{dt\over t}$$
(for $x\gt0$), then you can show that, for $a,b\gt0$, the linearity of integration and a simple substitution $t=au$ give
$$L(ab)=\int_1^{ab}{dt\over t}=\int_1^a{dt\over t}+\int_a^{ab}{dt\over t}=L(a)+\int_1^b{du\over u}=L(a)+L(b)$$
which is to say, the function $L$ has the defining property of a logarithm.  Showing that it is the "natural" logarithm function will depend on how you've defined the function $\ln x$.  In many (if not most) calculus texts, the integral is the definition of $\ln x$.  In some cases, though, the exponential function $e^x$, is defined first, say as 
$$e^x=\lim_{n\to\infty}\left(1+{x\over n}\right)^n$$
and $\ln x$ is defined as its inverse.  In that case it suffices to show that $L(e)=1$.  We can do this as follows, using the logarithmic nature of $L$ and fact that a function defined as an integral is continuous:
$$L(e)=\lim_{n\to\infty}L\left(\left(1+{1\over n}\right)^n\right)=\lim_{n\to\infty}\left(nL\left(1+{1\over n}\right)\right)=\lim_{n\to\infty}\left(n\int_1^{1+{1\over n}}{dt\over t} \right)$$
and ${1\over1+(1/n)}\le{1\over t}\le1$ for $1\le t\le1+{1\over n}$ tells us
$${1\over n}\cdot{1\over1+(1/n)}\le\int_1^{1+{1\over n}}{dt\over t}\le{1\over n}\cdot1$$
so that, by the squeeze theorem,
$$1=\lim_{n\to\infty}{1\over1+(1/n)}\le\lim_{n\to\infty}\left(n\int_1^{1+{1\over n}}{dt\over t} \right)\le\lim_{n\to\infty}1=1$$
which completes the proof that $L(e)=1$.
A: It is difficult to avoid circularity when addressing this question.
Anyway, you can remark that by a rescaling of the variable
$$L(x):=\int_1^x\frac{dx}x=\int_y^{xy}\frac{y\,dx}{y\,x}=\int_y^{xy}\frac{dx}x.$$
This can be written
$$L(xy)=L(x)+L(y),$$ as $L(1)$ is obviously $0$.
This gives you a famous functional equation.

From this, by induction
$$L(x^n)=nL(x),$$ then $$L(x^{n/m})=\frac nmL(x),$$ and by extension to the reals
$$L(x^y)=yL(x).$$ 
Now there must be a constant, let $e$, such that $L(e)=1,$
 then
$$L(e^x)=x$$
and  $L$ appears to be the inverse of an exponential.
