how do i find the coeffiecents A B (A/something)(B/Something) of the integral dx/(x-3)^2..............?

I don't know why you're looking to put the function in that particular format, but I would solve the problem using the method "u-substitution."

Let's say the variable u = x-3

Now we can say we are finding the integral of 1/(u^2), or (u^-2)

The integral of u^-2 is (u^-1/-1) = -u^-1 = -1/u

Sub x-3 back in for u to get the final answer of

-1/(x-3)

And don't forget the + C,

= -1/(x-3) + C

Remember that some questions will intentionally put dx in the numerator of the function to throw you off. Integral of dx/(x-3)^2 is the same thing as integral of 1/(x-3)^2 dx

All you need for this problem is to remember

d/dx ax^n = anx^(n-1) Since you're working 'back' the other way what would you need to have for a and n?

the integral is

-x^(-1) +C

Learning to do integrals is a lot like learning to factor trinomials. If you don't understand binomial multiplication, you'll never in a million years figure out factoring. In the same way, if you're having difficulties with problems this simple, you really, *really* should go back and make sure you fully understand differentiation.

HTH ☺

Doug