Asymptotic form of an integral to an power law decaying function

Question

$$ f(x)=\frac{1}{2}+\frac{1-x^2}{4x}\ln\left|\frac{1+x}{1-x}\right| $$

This function is not analytic at $x=1$. The plot is shown:

The integral is:

$$ I=\int_0^\infty g(x) \sin(2b rx) dx $$

where $g(x)$ is defined as:

$$ g(x)=\frac{x}{x^2+a^2 f(x)} $$

Both $a$ and $b$ are real positive. One can also see that $f(x)$ is even and thus $g(x)$ is odd.

In my book it is claimed that because the non-analytical behavior at $x=1$, at large $r$ we have a term behave as:

$$ \sim \frac{\cos(2br)}{r^2} $$

My question is how this asymptotic form is obtained

Add the numerical simulation

Setting $a=\sqrt{2},b=0.5$, we get the integrand:

f[x_, r_] := (x*
    Sin[r*x])/(x^2 + (1 + ((1 - x^2)*Log[Abs[(1 + x)/(1 - x)]])/(2*
         x)))

Then Integrate it numerically using mathematica:

lst = Table[{r, NIntegrate[f[x, r], {x, 0, \[Infinity]}]}, {r, 10, 50,
     pointSpacing}];

Find the fit curve using:

FindFit[lst, a Cos[b x + c]/x^2, {a, b, c}, x];

Then plot the fit curve(a,b,c is obtained above) and the data curve, they agrees very well:

Show[ListPlot[lst, PlotStyle -> Red], 
 Plot[(a Cos[b x + c])/
   x^2 /. {a -> 0.843455, b -> 1.00248, c -> 0.0594367}, {x, 1, 201}]]

Here is the numerical results: