To find the partial fraction decomposition, we begin with
#x/((x-6)(x-3)) = A/(x-6) + B/(x-3)#
Multiplying by #(x-6)(x-3)# gives us
#x = A(x-3)+B(x-6) = (A+B)x + (-3A - 6B)#
Matching respective coefficients on each side gives us the system of equations
#{(A + B = 1), (-3A -6B = 0):}#
Now all we need to do is solve for #A# and #B#. For example, using elimination,
#(-3A - 6B) + 3(A+B) = 0 + 3(1)#
#=> -3B = 3#
#=> B = -1#
Then, substituting
#A - 1 = 1#
#=> A = 2#
So the decomposition is
#x/((x-6)(x-3)) = 2/(x-6) - 1/(x-3)#
Now we can do the integral relatively easily.
#intx/((x-6)(x-3))dx = int(2/(x-6) - 1/(x-3))dx#
#int(2/(x-6) - 1/(x-3))dx = 2int(1/(x-6))dx - int(1/(x-3))dx#
A couple of simple #u# substitutions, together with #int(1/x)dx = ln|x|+C#
gives us
#2int(1/(x-6))dx = 2ln|x-6| + C#
and
#-int(1/(x-3))dx = -ln|x-3|+C#
So putting it all together, we get
#intx/((x-6)(x-3))dx = 2ln|x-6| - ln|x-3| + C#