The answer is no, even in the smooth case. Take for example:

$$
f(x) = \frac{2}{x} + \frac{\cos(\log(x))}{x}
$$

Alter it on a small neighborhood of $0$ in such a way that there is no singularity there, preserving smoothness (this will be irrelevant for the asymptotics). This function is decreasing and, for $t$ sufficiently large, we have

$$
\int_0^t f(x) dx = C + 2\log(t) + \sin(\log(t)) = 2\log(t) + o(\log(t))
$$

The monotone density theorem mentioned in the comments does not work in general if your r.h.s. is simply a slowly varying function (as any function asymptotic to $\log(t)$). You want your r.h.s. to be a **de Haan function**. The specific result you may want to use is Theorem 3.6.8 here:

<cite authors="Bingham, N. H.; Goldie, C. M.; Teugels, J. L.">_Bingham, N. H.; Goldie, C. M.; Teugels, J. L._, Regular variation., Encyclopedia of Mathematics and its Applications, 27. Cambridge etc.: Cambridge University Press. 512 p. £ 20.00/pbk (1989). [ZBL0667.26003](https://zbmath.org/?q=an:0667.26003).</cite>