Who discovered the non-obvious $\int \frac 1x dx=\ln(x)+c$ ? Were power series involved? The series look similar on opposite sides of 1:
$$ \frac 1x =\sum_{n=0}^\infty (-1+x)^n \text{ for } |x-1|<1 $$
$$ \ln(x) = \ln(-1+x)-\sum_{k=1}^\infty \frac{(-1)^k}{k(-1+x)^k} \text{ for } |x-1|>1 $$
$$ \ln(x) = -\sum_{k=1}^{\infty}\frac{(-1)^k(-1+x)^k}k \text{ for } |x-1|<1 $$
If: $\frac ab = \frac cd$ then $\int_a^b\frac 1x dx=\int_c^d\frac 1x dx$
If $\frac ab = \frac cd$ and $a=1$ and $c=1$ then $\frac 1b \not= \frac 1d$, but $\frac 1b + \frac 1d = \frac {1}{bd}$
This is the same as saying what RGB did: $$\int_{1}^{bd}\frac{\text{d}t}{t}=\int_{1}^{b}\frac{\text{d}t}{t}+\int_{1}^{d}\frac{\text{d}t}{t}$$
$$f(bd)=f(b)+f(d)$$
$$ A^b A^d=A^{b+d}$$
Thus a logarithmic relationship was implied and A was found to be e.
$$ \int_1^b \frac 1x dx = \ln(b)$$ $$ \int_1^d \frac 1x dx = \ln(d)$$ $$ \int_1^{bd} \frac 1x dx = \ln(b)+\ln(d)$$
Source
The Hyperbolic Angle "$u$" is defined in relation to the function $\frac 1x$:
The magnitude of the hyperbolic angle "$u$" is the area of the corresponding hyperbolic sector (red above, yellow below) which is ln x since it is defined as the integral of the projection onto the axis from a=1 or y=x to b: $\int_1^b \frac 1x dx=\ln(b)$
While power series were not involved in the discovery of the natural logarithm, they can be used to show the relationship between $\frac 1x$ and e:
If a function is its own derivative then: $ y=f(x), \frac{dy}{dx}=y$
Such a function is a series: $ y=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots$
$$\frac{dy}{dx}=0+1+\frac{2x}{2!}+\frac{3x^2}{3!}+\frac{4x^3}{4!}\dots=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+ \dots$$
If we differentiate $y^p$: $ \frac{d(y^p)}{dx}=py^{p-1}\frac{dy}{dx}$
Since $ \frac{dy}{dx}=y$ it follows: $ \frac{d(y^p)}{dx}=py^{p-1}y=py^p$
Given: $y_1=1+z+\frac{(z)^2}{2!}+ \frac{(z)^3}{3!}+\dots$
$$\frac{dy_1}{dz}= 1+z+\frac{(z)^2}{2!}+ \frac{(z)^3}{3!}+\dots=y_1$$
Let $z=ax$,$\frac{dy_1}{d(ax)}=y_1$ $\frac{dy_1}{d(x)}=ay_1$
$$y_1=y^p= 1+px+\frac{(px)^2}{2!}+ \frac{(px)^3}{3!}+\dots=y_1$$
If we let $p=\frac 1x$ then, $y^{\frac 1x}=1+1+\frac{(1)^2}{2!}+ \frac{(1)^3}{3!}+\dots=e$
And Hence replacing $e^1$ with $e^x$ yeilds:
$e^x=1+x+\frac{(x)^2}{2!}+ \frac{(x)^3}{3!}+\dots$
This all lead to the development of the Slide Rule as well as hyperbolic trig functions.