1. Cambio de variable:
Sea $u = \sqrt{x - 1} \Rightarrow x = u^2 + 1$, $dx = 2u \, du$.
El denominador se transforma: $x^2 - 2x + 2 = (x-1)^2 + 1 = (u^2)^2 + 1 = u^4 + 1$.

2. Integración:
$$ \int \frac{(u^2 + 1) 2u \, du}{(u^4 + 1) u} = 2 \int \frac{u^2 + 1}{u^4 + 1} du $$
Dividimos por $u^2$:
$$ 2 \int \frac{1 + \frac{1}{u^2}}{u^2 + \frac{1}{u^2}} du $$
Sea $v = u - \frac{1}{u}$, entonces $dv = (1 + \frac{1}{u^2}) du$ y $v^2 + 2 = u^2 + \frac{1}{u^2}$.
$$ 2 \int \frac{dv}{v^2 + 2} = \frac{2}{\sqrt{2}} \arctan\left( \frac{v}{\sqrt{2}} \right) + C = \sqrt{2} \arctan\left( \frac{u^2 - 1}{u\sqrt{2}} \right) + C $$

3. Resultado:
Sustituyendo $u = \sqrt{x-1}$:
$$ \boxed{\sqrt{2} \arctan\left( \frac{x - 2}{\sqrt{2(x-1)}} \right) + C} $$