$\int \dfrac{x}{(x-1)(x-2)^2} d x=a \log \left|\dfrac{x-1}{x-2}\right|+\dfrac{b}{(x-2)}+c$ then

Let the integrand be $\dfrac{x}{(x-1)(x-2)^2} = \dfrac{A}{x-1} + \dfrac{B}{x-2} + \dfrac{C}{(x-2)^2}$.

Multiplying by $(x-1)(x-2)^2$, we get $x = A(x-2)^2 + B(x-1)(x-2) + C(x-1)$.

Setting $x=1$, we get $1 = A(1-2)^2$, so $A=1$.

Setting $x=2$, we get $2 = C(2-1)$, so $C=2$.

Comparing the coefficients of $x^2$, we get $0 = A + B$, so $B = -A = -1$.

Thus, the integral is $\int \left( \dfrac{1}{x-1} - \dfrac{1}{x-2} + \dfrac{2}{(x-2)^2} \right) dx$.

Evaluating the integral, we get $\ln|x-1| - \ln|x-2| - \dfrac{2}{x-2} + c = \ln\left|\dfrac{x-1}{x-2}\right| - \dfrac{2}{x-2} + c$.

Comparing this with the given form $a \log \left|\dfrac{x-1}{x-2}\right| + \dfrac{b}{x-2} + c$, we identify $a=1$ and $b=-2$.

Answer: $a=1, b=-2$