jmc

Multiplying both sides by $(x-2{)}^3(x-1)$ gives us $$4x^3-20x^2+37x-25=A(x-2{)}^3+B(x-1)+C(x-1)(x-2)+D(x-1)(x-2{)}^2.$$Setting $x=2$ gives $4(8)-20(4)+74-25=B$. Evaluating the expression on the left gives us $B=1$.

Setting $x=1$ gives $4-20+37-25=A(-1{)}^3$, and so $A=4$.

We still need to find $C$ and $D$. By choosing 2 new values for $x$, we can get two equations which we can solve for $C$ and $D$. We can pick convenient values to make our work easier.

When $x=0$, we get $$-25=4(-2{)}^3+(-1)+C(-1)(-2)+D(-1)(-2{)}^2$$which simplifies to $$2C-D=8.$$When $x=-1$, we get $$4(-1{)}^3-20(-1{)}^2+37(-1)-25=4(-3{)}^3+(-2)+C(-2)(-3)+D(-2)(-3{)}^2$$which simplifies to $$C-3D=4.$$We can multiply this equation by $2$ and subtract it from the previous one to get $-D+6D=8-2\cdot 4=0$ and hence $D=0$. Then $2C=8$ and $C=4$. Therefore $(A,B,C,D)=\boxed{(4,1,4,0)}.$