Question:

Let $x = 1/t$, then $dx = -1/t^2 dt$. The integral becomes $\int \frac{(1/t - 1) (-1/t^2) dt}{(1/t^2) \sqrt{2/t^2 - 2/t + 1}} = \int \frac{-(1-t)/t dt}{\sqrt{(2-2t+t^2)/t^2}} = \int \frac{-(1-t) dt}{\sqrt{t^2-2t+2}} = \int \frac{(t-1) dt}{\sqrt{(t-1)^2+1}}$. Let $u = t-1$, then $du = dt$. The integral is $\int \frac{u du}{\sqrt{u^2+1}} = \sqrt{u^2+1} + C = \sqrt{(t-1)^2+1} + C = \sqrt{(1/x - 1)^2 + 1} + C = \sqrt{\frac{(1-x)^2 + x^2}{x^2}} + C = \frac{\sqrt{2x^2-2x+1}}{x} + C$. Comparing with $\frac{\sqrt{f(x)}}{g(x)}$, we get $f(x) = 2x^2-2x+1$ and $g(x) = x$.