The problem says to use the fact that
$g(x) = \int_0^\infty f(t) e^{-xt}$
is its own Fourier sine transform if $f(x)$ is its own cos transform.
My working so far:
$F_c(\int_0^\infty f(t) \frac{1}{t+x} dt) = \int_0^\infty \int_0^\infty cos(sx) f(t) \frac{1}{t+x} dt dx$
Now reverse the order of integration
$F_c(\int_0^\infty f(t) \frac{1}{t+x} dt) = \int_0^\infty \int_0^\infty cos(sx) f(t) \frac{1}{t+x} dx dt$
Use the fact that the Laplace transform of $e^{-ap}$ is $\frac{1}{a+p}$ to substitute for $\frac{1}{t+x}$:
$F_c(\int_0^\infty f(t) \frac{1}{t+x} dt) =\int_0^\infty \int_0^\infty cos(sx) f(t) \int_0^\infty e^{-ty}e^{-xy} dy dx dt$
I think the next step would be to change the order of integration and somehow work with the function $f(t)e^{ty}$ since that appears in $g(x)$, but I'm not sure what to do.