<s>[INST] 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. 
 [/INST] $$\frac{3x-2}{(x-2)^2} = \frac{A}{x-2} + \frac{B}{(x-2)^2}$$
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
$$\frac{3x-2}{(x-2)^2} = \frac{3}{x-2} + \frac{4}{(x-2)^2}$$
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