Using the Riemann Integral and the fact that the indefinite integral of $\frac{1}{x}$ is $\ln|x| + C$ I get
$$\int_a^b \frac{1}{x}dx = \lim_{n \to \infty} \sum_{i=1} ^{n} \frac{1}{a+\Delta xi}\Delta x=(\ln|b|+C)-(\ln|a|+C)=\ln|b|-\ln|a|$$
where $\Delta x =\frac{b-a}{n}$
Taking $\Delta x$ out of the numerator and denominator
$$\lim_{n \to \infty} \sum_{i=1} ^{n} \frac{1}{\frac{a}{\Delta x} + i}\frac{\Delta x}{\Delta x}=\ln|b|-\ln|a|$$
$$\lim_{n \to \infty} \sum_{i=1} ^{n} \frac{1}{\frac{a}{\Delta x} + i}=\ln|b|-\ln|a|$$
Replacing $\Delta x$ with $\frac{b-a}{n}$
$$\lim_{n \to \infty} \sum_{i=1} ^{n} \frac{1}{\frac{a}{\frac{b-a}{n}} + i}=\ln|b|-\ln|a|$$
$$\lim_{n \to \infty} \sum_{i=1} ^{n} \frac{1}{\frac{an}{b-a} + i}=\ln|b|-\ln|a|$$
Substituting $a=1$
$$\lim_{n \to \infty} \sum_{i=1} ^{n} \frac{1}{\frac{n}{b-1} + i}=\ln|b|-\ln|1|=\ln|b|$$
Finally, replacing $b=x$
$$\ln|x| = \lim_{n \to \infty} \sum_{i=1} ^{n} \frac{1}{i + \frac{n}{x-1}}$$
As far as I'm aware there is no known series of $\ln|x|$ where none of the terms have exponents. Could this particular series be of any use in the approximation of natural logarithm values, or any other application of the natural logarithm?