$\int \frac{x d x}{(x-1)(x-2)}=$

Decompose the integrand using partial fractions:

$\frac{x}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2}$

Clearing denominators yields $x = A(x-2) + B(x-1)$.

At $x = 1$: $1 = A(-1)$, so $A = -1$.
At $x = 2$: $2 = B(1)$, so $B = 2$.

The decomposition becomes $\frac{-1}{x-1} + \frac{2}{x-2}$.

Integrating term by term:

$-\int \frac{1}{x-1}\,dx + 2\int \frac{1}{x-2}\,dx = -\ln|x-1| + 2\ln|x-2| + C$

Applying logarithmic identities:

$\ln|(x-2)^2| - \ln|x-1| + C = \ln\left|\frac{(x-2)^2}{x-1}\right| + C$

This expression corresponds to option D.

Final answer: $\boxed{D}$