M missyc8 New member Joined Sep 7, 2009 Messages 13 Sep 16, 2009 #1 integral (3x^2 - 10/ x^2 -4x+ 4) dx bottom = (x-2) (x-2) A/(x-2) + B/ (x-2)^2 where do i go from here? i do not understand...thanks!
integral (3x^2 - 10/ x^2 -4x+ 4) dx bottom = (x-2) (x-2) A/(x-2) + B/ (x-2)^2 where do i go from here? i do not understand...thanks!
D Deleted member 4993 Guest Sep 16, 2009 #2 missyc8 said: integral (3x^2 - 10/ x^2 -4x+ 4) dx bottom = (x-2) (x-2) A/(x-2) + B/ (x-2)^2 where do i go from here? i do not understand...thanks! Click to expand... First you need to write it in a way that numerator is lower degree polynomial than denominator (right now those are both 2nd degree polynomial) \(\displaystyle \frac{3x^2 - 10}{x^2 - 4x + 4} \, = \, 3 \, + \, 2\cdot \frac{6x - 11}{x^2 - 4x + 4}\) Then \(\displaystyle \frac{6x - 11}{x^2 - 4x + 4} \, = \, \frac{A}{x-2} \, + \, \frac{B}{(x-2)^2}\) \(\displaystyle \frac{6x - 11}{x^2 - 4x + 4} \, = \, \frac{A(x-2) + B}{(x-2)^2}\) Now solve for A & B by equating the numerators
missyc8 said: integral (3x^2 - 10/ x^2 -4x+ 4) dx bottom = (x-2) (x-2) A/(x-2) + B/ (x-2)^2 where do i go from here? i do not understand...thanks! Click to expand... First you need to write it in a way that numerator is lower degree polynomial than denominator (right now those are both 2nd degree polynomial) \(\displaystyle \frac{3x^2 - 10}{x^2 - 4x + 4} \, = \, 3 \, + \, 2\cdot \frac{6x - 11}{x^2 - 4x + 4}\) Then \(\displaystyle \frac{6x - 11}{x^2 - 4x + 4} \, = \, \frac{A}{x-2} \, + \, \frac{B}{(x-2)^2}\) \(\displaystyle \frac{6x - 11}{x^2 - 4x + 4} \, = \, \frac{A(x-2) + B}{(x-2)^2}\) Now solve for A & B by equating the numerators
