I know that in general (though I don't know the EXACT conditions on $f$)
$$\int_0^\infty f(x)\text{d}x = \int_0^\infty x^{-2}f(x^{-1})\text{d}x$$
so I am wondering if it obviously follows that the following integral diverges
$$\int_0^\infty \frac{\ln(1+x)}{x}\text{d}x $$
my idea was that by the above trick (if this integral exists), this integral must also equal
$$\int_0^\infty \frac{\ln(1+\frac{1}{x})}{x}\text{d}x$$
however, I'm not sure how to finish the argument. It seems obvious if the integral began from $1$ but since it passes through $(0,1)$ first $x^{-1}$ in some sense is the same as $(1,\infty)$ for $x$.... so that the terms might somehow balance out. Anyway, just curious if my argument can meaningfully be completed. (I'm aware this integral diverges but want to see if my proof works or can be made to work)
Also, I would appreciate any clarification on the conditions for $f$ so that the first equation holds.