I was playing around with some calculus, attempting to prove for myself: $$\int_b^a{\frac{1}{x}} dx= \ln(a) -\ln(b)$$ However I came to a halt when the method I was using required me to prove the following: $$\lim_{h\to\infty}\bigg(\sum_{r=0}^{\infty}\frac{(a-b)}{r(a-b)+bh}\bigg) \equiv \ln(a) - \ln(b)$$ I am confident that they are equivalent, after testing a bunch of random values.
I don't want to use the fact that differentiation and integration are inverse operations. I am curious about how one would go about proving this (since I'm still at high school, so I haven't yet learned how to deal with difficult problems such as these)!
I think using a series to get to the logarithm won't get you very far.
In fact we'd rather calculate such series as a Riemann sum leading to integration of $\frac 1r$ and giving $\ln$ as a result.
Instead it is possible to play with the properties of the integral.
Let set $\displaystyle f(x)=\int_1^x \dfrac{dt}t$
For instance substitute $t=e^u$ then $$f(e^\alpha)=\int_1^{e^\alpha}\dfrac{dt}{t}=\int_0^\alpha \dfrac{e^u\mathop{du}}{e^u}=\int_0^\alpha du=\alpha$$
First by substituting $u=\frac 1t$ then $$f(\frac 1x)=\int_1^{\frac 1x}\dfrac{dt}{t}=\int_1^x \dfrac{-u\mathop{du}}{u^2}=-\int_1^x du=-f(x)$$
Then by substituting $t=bu$ then $$f(ab)=\int_1^{ab}\dfrac{dt}{t}=\int_{\frac 1b}^a \dfrac{b\mathop{du}}{bu}=\int_{\frac 1b}^a \dfrac{du}{u}=f(a)-f(\frac 1b)=f(a)+f(b)$$