The numerator is what you get by differentiating the contents of the brackets (save for a factor of 2). So "guess" that the answer is something like #(x^2+6x-5)^-1#, and differentiate this back to get #-(2x+6)(x^2+6x-5)^-2#. So this is the original integrand times #-2#, so the guess was two times too high and the wrong sign. So we re-guess as #-1/2(x^2+6x-5)^-1# which checks out upon differentiating.
Alternatively, if you think is too "guessy' you could just substitute #u=x^2+6x-5# and see what turns up:
#{du}/{dx}=2x+6#,
or #{dx}/{du}=1/(2(x+3)# so the integral is:
#int (x+3)/(x^2+6x-5)^2 {dx}/{du}du#
#=int cancel(x+3)/u^2 * 1/(2(cancel(x+3)))du#
#=(1/2)int1/u^2du#
#=-(1/2).(1/u)+C#
#=-1/(2(x^2+6x-5))+C#
The key step is that all the #x#'s cancelled out in the odd mixed-up expression with both #x#'s and #u#'s. This happened only because of the connection between the numerator and the denominator.
