How to prove a function defined by a definite integral is unbounded without computing the antiderivative.

I am trying to show the following function, $f:\mathbb{R}\rightarrow\mathbb{R}$, defined by $$f(x) = \int_{1}^{x} \frac{1}{t} dt$$

increases without bound as $x$ increases without bound and decreases without bound as $x$ approaches $0$ only using basic properties of the definite integral (i.e. Fundamental Theorem of Calculus, sum rule, constant multiplication rule, domain sum rule). Of course, this integral must diverge since the log function is unbounded, but I am approaching the problem without exploiting this fact. Using the fact that the integral is the difference in the antiderivative of $1/t$ (FTC) is permissible, however. I am wondering if we can show this function satisfies the same properties as the log function and then use those properties to show the desired unboundedness. Any thoughts on this matter would be much appreciated.

\begin{align} \int_{2^n}^{2^{n+1}}\frac{dt}t &=\int_1^2\frac{2^ndx}{2^nx}\\ (t&\mapsto2^nx)\\ &=\int_1^2\frac{dx}x\\ &:=\alpha>0 \end{align}

Thus: \begin{align} \int_1^\infty\frac{dt}t&=\\ \int_1^2\frac{dt}t+&\int_2^4\frac{dt}t+\int_4^8\frac{dt}t+\dotsb\\ &=\alpha+\alpha+\alpha+\dotsb\\ &=\infty \end{align} If you want to be fancy, you could say we're using the Archimedean property of the reals in that last bit.

Note that $$f(2^{n+1}) - f(2^n) = \int_{2^n}^{2^{n+1}} \dfrac{dt}{t} \ge \dfrac{2^{n+1}-2^n}{2^{n+1}} = \dfrac{1}{2}$$