Which of these is a way of expressing the natural log of x?

The indefinite integral of $\frac{1}{x}$ isn't some anomaly. The integrating $x^n$ for powers close but not equal to $-1$ is indeed approximately close to the natural logarithm with a "constant" to determine the constant, consider two points that are true for a logarithm in any base; the y-intercept and the root.

$\int x^n dx=\frac{x^{n+1}}{n+1}$ +c

Take the limit as $n$ approaches $-1$

$\int x^{-1} dx=\lim_{n \to -1}\frac{x^{n+1}}{n+1}+c$

Change the left hand side to the natural logarithm.

$\ln{x}+k=\lim_{n \to -1}\frac{x^{n+1}}{n+1}+c$

Let $k=0$ and solve for $c$

$\ln{x}=\lim_{n \to -1}\frac{x^{n+1}}{n+1}+c$

Let $x=1$ because the logarithm can be expressed exactly there.

$\ln{1}=\lim_{n \to -1}\frac{1^{n+1}}{n+1}+c$

$0=\lim_{n \to -1}\frac{1^{n+1}}{n+1}+c$

$\therefore c=\lim_{n \to -1}\frac{-1}{n+1}$

Combine fractions because they share a common denominator

$\therefore \ln{x}=\lim_{n \to -1}\frac{x^{n+1}-1}{n+1}$