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?

We can use the following method
$$\frac1{(x-1)^2(x-2)^2}=\frac{\{(x-1)-(x-2)\}^2}{(x-1)^2(x-2)^2}=\frac1{(x-2)^2}+\frac1{(x-1)^2}-2\frac1{(x-1)(x-2)}$$
$$\frac1{(x-1)(x-2)}=\frac{(x-1)-(x-2)}{(x-1)(x-2)}=\frac1{x-2}-\frac1{x-1}$$

Alternatively, 
Put $x-2=y$ to ease of calculation
$$1=A(y+1)y^2+By^2+C(y+1)^2y+D(y+1)^2$$
$$\implies 1=D+y(C+2D)+y^2(A+B+2C+2D)+y^3(A+C)$$
As this is an identity, we can compare the coefficients of the different powers of $y$
Comparing the coefficients of $y^0,D=1$
Comparing the coefficients of $y,C+2D=0\implies C=-2D=-2$
Comparing the coefficients of $y^3,A+C=0\implies A=-C=2$
Comparing the coefficients of $y^2,A+B+2C+2D=0$
May I leave this for you to find $B?$
