H hgaon001 New member Joined May 17, 2009 Messages 39 Jul 30, 2009 #1 I have the integral of [(15x-3)/(3x-1)^2] after i put in the a and the b over (3x-1) and (3x-1)^2 i get stuck because by pluggin in 1/3 it cancels both of them out
I have the integral of [(15x-3)/(3x-1)^2] after i put in the a and the b over (3x-1) and (3x-1)^2 i get stuck because by pluggin in 1/3 it cancels both of them out
H hgaon001 New member Joined May 17, 2009 Messages 39 Jul 30, 2009 #2 well i end up with 15x-3=A(3x-1)^2 + B(3x-1) but when pluggin in x=1/3 to get rid of one of them both will cancel out..
well i end up with 15x-3=A(3x-1)^2 + B(3x-1) but when pluggin in x=1/3 to get rid of one of them both will cancel out..
