Can you work out the integral of $1/x$ by integration? The way you learn to do this kind of problem in highschool is:

*

*"we know the solution is $\ln(x)$, because if we take the derivative
of $\ln(x)$ it's $1/x$"

but i was wondering if there was a way to get this solution by working out the integral instead of taking it as an antiderivative.
In my highschool the way we defined $\ln(x)$ is as the inverse function of $e^x$.
 A: We can use by substitution
$$x=e^u \implies dx = e^u du$$
and then
$$\int \frac1x dx= \int \frac1{e^u}e^u du =\int du = u+C$$
and then use the definition for the inverse function of $e^x$.

As an alternative, we can use the "derivative of inverse function" rule
$$(f^{-1})'(y) = \frac{1}{f'(x)}$$
and since $(e^x)'=e^x$, indicating with $\ln x$ the inverse (i.e. $y=e^x \iff x=\ln y$) we have
$$(\ln y)' = \frac1{e^x}=\frac1y$$

Refer also to the related

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*Intuition for the derivative of the exponential function
A: One way to approach it is defining the integral an dthen work out properties of that function.  To let's define $L:\Bbb R^+\to\Bbb R$ as
$$L(x)=\int\limits_1^x\frac1tdt\tag 1$$
One observation is that when scaling the integrand by a factor of $a$, then we still get the same function $L$: So observe what happens when we substitute $u=at$, $du=a\,dt$:
$$L(x)=\int\limits_a^{ax}\frac1udu\tag 2$$
So only the bounds of integration did change, but the integrand is unaltered! The right side of $(2)$ can be completed to an integral starting at 1 by supplementing with a like integral from $1$ to $a$:
$$L(a)+L(x)
=\int\limits_1^{a}\frac1tdt + \int\limits_a^{ax}\frac1udu
=\int\limits_1^{ax}\frac1tdt=L(ax)\tag 4$$
Hence the function $L$ satisfies the functional equation $$L(xy)=L(x)+L(y)\tag 5$$ and in addition $L$ is continuous$^1$, which shows that $L$ must be some logarithmic function.  One way to see this is to know that $(5)$ is a functional equation satisfied by logarithms.  An other way to see it, is that the inverse $E$ of $L$, which clearly exists$^2$, must obey the functional equation
$$E(x)E(y)=E(x+y)\tag 6$$
which reveals $E$ to be an exponential function of some sort. Now we also have the obvious
$$L(1)=0 \quad\text{ and }\quad L'(1)=1$$
which translates to the inverse $E$ as
$$E(0)=1 \quad\text{ and }\quad E'(0)=1$$
To see this, just reflect $L$ at the line $x=y$ to get $E$, or differentiate $E(L(x)) = x$ to get $L'(x)E'(L(x)) = 1$ and then set  $x=1$.

$^1L$ is continuous because it is the integral of a smooth function.
$^2L$ is the integral of a function that's positive throughout, thus $L$ is strictly increasing and thus injective.
