The following equation seems extremely simple:
$$ g(b)=\int_1^{\frac{1}{b}} \frac{X f(X)}{\sqrt{1-b X}} \, dX $$
But how to solve it, that is, restore the function $f(X)$ from the known function $g(b)$? Maybe this is some well-known integral transformation, which I was not told about at the university? Additionally, we can assume that $f(1)=0$.
This is not a complete solution, just two ideas which are too large for a comment
let us define helper function $h$, so that: $$g(b)=h(1/b), $$
now we have $$h(t) = \int_1^{t} \frac{X f(X)}{\sqrt{1- X/t}} \, dX $$ and use chain rule with fundamental theorem
$h'(t)=\frac{tf(t)}{\sqrt{1-t/t}}$
since we get division by zero corollary is not enough and we may have to go to https://en.wikipedia.org/wiki/Leibniz_integral_rule#General_form:_Differentiation_under_the_integral_sign
In particular
$$\frac{d}{dx}\left(\int_a^{x} f(x,t)dt\right) = f(x,x)+\int_a^{x} \frac{d}{dx}f(x,t) dt$$
or if we are comfortable with Fourier transforms we can possibly use convolution theorem with Fourier transform of
$\frac{1}{\sqrt{1-t}}$