Starting with $g(x)=\int_{1/x}^1 g(kx)(2-2k)dk+1,$ after integration by parts of the RHS I get the following DE for the second antiderivative of $g(x),$ say $f(x)$:
$x^2f''(x)+(2x-2)f'(1)-2(f(x)-f(1))-1=0.$
I'm not really sure how to approach this - for instance, if I pretend $f(1)$ and $f'(1)$ are merely constants, I can spot the differential operator
$x^2f''(x)-2f(x)$
has eigenfunctions $1/x$ and $x^2$. However, the usual plan of using these homogeneous solutions to find the general solution seems not to work, as my homogeneous solutions (and their first derivatives) don't vanish at $x=1$.
Is there perhaps a better way of attacking the original integral equation? It's also worth mentioning I have a BC - namely $g(1)=1.$
$$x^2f''(x)+(2x-2)f'(1)-2(f(x)-f(1))-1=0.$$ $$x^2f''(x)-2f(x)=Ax+B$$ $$x^2f''(x)\color{red}{+2xf'-2xf'}-2f(x)=Ax+B$$ $$(x^2f'(x))'-2(xf(x))'=Ax+B$$ $$(x^2f'(x)-2xf(x))'=Ax+B$$ Integrate both sides to reduce the order of the DE. Or you can rexrite the DE as: $$\left ( x^4 \left(\dfrac {f(x)}{x^2}\right)'\right)'= Ax + B$$ Integrate twice.