Partial fraction integration $\int \frac{dx}{(x-1)^2 (x-2)^2}$ $$\int \frac{dx}{(x-1)^2 (x-2)^2} = \int \frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x-2}+\frac{D}{(x-2)^2}\,dx$$
I use the cover up method to find that B = 1 and so is C. From here I know that the cover up method won't really work and I have to plug in values for x but that won't really work either because I have two unknowns. How do I use the coverup method?

$$
\frac1{(x-1)^2(x-2)^2}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x-2}+\frac{D}{(x-2)^2}\tag{1}
$$
Multiply both sides by $(x-1)^2$ and evaluate at $x=1$: $B=1$
Multiply both sides by $(x-2)^2$ and evaluate at $x=2$: $D=1$
$$
\frac{1-(x-1)^2-(x-2)^2}{(x-1)^2(x-2)^2}=\frac{A}{x-1}+\frac{C}{x-2}\tag{2}
$$
Multiply both sides by $x-1$ and use L'Hospital at $x=1$: $A=2$
Multiply both sides by $x-2$ and use L'Hospital at $x=2$: $C=-2$
$$
\frac1{(x-1)^2(x-2)^2}=\frac{2}{x-1}+\frac{1}{(x-1)^2}+\frac{-2}{x-2}+\frac{1}{(x-2)^2}\tag{3}
$$

Details of $(2)$
It is pretty simple to subtract off the parts we got from $(1)$:
$$
\frac1{(x-1)^2(x-2)^2}-\frac1{(x-1)^2}-\frac1{(x-2)^2}
=\frac{1-(x-1)^2-(x-2)^2}{(x-1)^2(x-2)^2}\tag{4}
$$
$(4)$ gives us $(2)$.
In the next part, we can use a small bit of simplification. Suppose we have $(x-a)^nP(x)$ in the denominator, and we want to look at $x\to a$. We are going to need to apply L'Hospital $n$ times. This will give $n!P(a)$ in the denominator.
Multiply $(2)$ by $(x-1)$ and let $x\to1$. The denominator is $(x-1)(x-2)^2$, so we apply L'Hospital once.
Take one derivative of the denominator and evaluate at $\color{#C00000}{x=1}$: $1!(\color{#C00000}{1}-2)^2=1$.
Take one derivative of the numerator and evaluate at $\color{#C00000}{x=1}$: $-2(\color{#C00000}{1}-1)-2(\color{#C00000}{1}-2)=2$
Thus, $A=\frac21=2$
Multiply $(2)$ by $(x-2)$ and let $x\to2$. The denominator is $(x-1)^2(x-2)$, so we apply L'Hospital once.
Take one derivative of the denominator and evaluate at $\color{#C00000}{x=2}$: $1!(\color{#C00000}{2}-1)^2=1$.
Take one derivative of the numerator and evaluate at $\color{#C00000}{x=2}$: $-2(\color{#C00000}{2}-1)-2(\color{#C00000}{2}-2)=-2$
Thus, $C=\frac{-2}1=-2$
