This integrates faster if we just integrate it as indefinite and then use FTC. Assuming the integral is proper, which I did not check.
ClearAll[x, y, k];
num = 2 x^2 - 6 x y + 3 + x^2 y^2;
den = x^2 (x - 2 y)^2;
int = Integrate[Sqrt[y^2 - 1] num/den, y]
upper = Limit[int, y -> (x^2 + 1)/(2 x)];
lower = Limit[int, y -> 1];
(upper - lower) // Simplify
f[x_] := Evaluate[upper - lower];
Plot[Chop@f[x], {x, 0, 2}, Frame -> True,
FrameLabel -> {{"integral", None}, {"x", "integral over x"}},
GridLines -> Automatic, GridLinesStyle -> LightGray,
Exclusions -> None, BaseStyle -> 14, PlotStyle -> Red]
