\int_1^x t^{-1 + \epsilon} dt
For \epsilon greater than 0. Obviously this would be:
\frac{x^{\epsilon} - 1}{\epsilon}
And you can see by applying l'hopitals that this goes to ln(x) as \epsilon goes to 0.
Now if you want to get more technical, and not assume anything, you can see that the derivative of a^x is:
\lim_{h \rightarrow 0} \frac{a^{x+h} - a^x}{h} = a^x \lim_{h \rightarrow 0} \frac{a^h - 1}{h}
Now if the limit exists, which you can see it clearly does by looking at the graph, this means that the derivative is some function of a times a^x. You can easily check that for a less than one, this function of a is negative, its equal to 0 at one, and its greater than 0 for a greater than one, and continues getting bigger and bigger. For some a, it will be exactly equal to one, and we'll call this e. So the derivative of e^x is just e^x. The inverse of exponentiation is logarithm, and ln(x) is defined as the log base e. Then the derivative of a^x is the derivative of e^(ln(a)*x) which you can see from the chain rule is ln(a)*e^(ln(a)*x), or just ln(a)*a^x. So the function of a was just ln(a). This can be used in the limit above:
\lim_{\epsilon \rightarrow 0} \frac{x^{\epsilon} - 1}{\epsilon} = \lim_{\epsilon \rightarrow 0} \frac{ln(x) \cdot x^{\epsilon}}{1} = ln(x)
Actually, you don't even have to use l'hopitals at this point, because if you look carefully you'll see this limit and the funtion of a are indentical, just with a replaced by x and h replaced by \epsilon.