Solve this integral by using cover up method in partial fractions new user here. I am trying to decompose the following fraction so that it can be solved easily. 
$$\frac{3x-2}{(x-2)^2} = \frac{A}{x-2} + \frac{B}{(x-2)^2}$$
General formula: 
$$\frac{px+q}{(x-a)^2} = \frac{A}{x-a} + \frac{B}{(x-a)^2}$$
I am able to find the value of A by simply equating it to p (the coefficient of x) but I'm having trouble determining B. On substituting x=2 in either of the denominators, $$(x-2)$$ or $$(x-2)^2$$ They become 0 resulting in an indeterminant form of the fraction. Please help me out. 
 A: The usual thing is to clear all denominators, resulting in
$$px+q=A(x-a)+B.$$
Simply plugging in $x=a$ will tell you what $B$ is; to find $A$ you can either differentiate both sides and set $x=a$ or you can simply equate coefficients to begin with. Or you can use another test point other than $a$.
(To be a little bit pedantic, the equation is only known to hold for $x \neq a$, since the original equation only held for $x \neq a$, so the procedure being used to find $B$ is actually sending $x \to a$. But the result is the same.)
A: The fast way to find $A$ and $B$ is this:
Remove the denominators multiplying both sides by $(x-2)^2$; you obtain
$$3x-2=A(x-2)+B.$$
Now, setting $x=2$ yields instantly $\;\color{red}4=A\cdot 0+B\color{red}{=B}$.
Next, identifying the leading coefficients of both sides yields $\color{red}{3=A}$, whence the decomposition
$$\frac{3x-2}{(x-2)^2}=\frac 3{x-2}+\frac 4{(x-2)^2}.$$
Note: of course, one may write $\; 3x-2=3(x-2)+6-2=3(x-2)+4$ and split the fraction in two, but the above method is more general.
A: $$\frac{3x-2}{(x-2)^2} = \frac{A}{x-2} + \frac{B}{(x-2)^2}$$
If and only if
$3x-2 = A(x-2) + B$
If you take $x = 2 \implies B = 4$
Then $3x-2 = A(x-2) + 4$
If you take $x = 0 \implies A = 3$
Then
$$\frac{3x-2}{(x-2)^2} = \frac{3}{x-2} + \frac{4}{(x-2)^2}$$
