View Full Version : integration partial fractions
1. The problem statement, all variables and given/known data
integrate (4x^2 + 3x + 6)/x^2 (x+2) dx
2. Relevant equations
don't have sorry..
3. The attempt at a solution
firstly = A/x + B/x^2 + C /x+2 , = A(x^2)(x+2) + B(x)(x+2) + C(x)(x^2) equating with the 4x^2 + 3x + 6,then i integrate it,but my ans turn out to be wrong..so could you please rectify my mistakes..thanks
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
tiny-tim
Jan1-09, 04:23 AM
Hi noobie!! :smile:
(4x^2 + 3x + 6)/x^2 (x+2) dx
…
= A/x + B/x^2 + C /x+2 , = A(x^2)(x+2) + B(x)(x+2) + C(x)(x^2) equating with the 4x^2 + 3x + 6
Nooo … too many x's! :wink:
get some sleep :zzz:
then try again! :smile:
Try to obtain the numerator as a derivative of the denominator. And then the extra term which you get try to break it in simple parts.
Hi Noobie,
1. The problem statement, all variables and given/known data
integrate (4x^2 + 3x + 6)/(x^2 (x+2)) dx
2. Relevant equations
don't have sorry..
3. The attempt at a solution
firstly = A/x + B/x^2 + C /x+2 , = A(x^2)(x+2) + B(x)(x+2) + C(x)(x^2) equating with the 4x^2 + 3x + 6,then i integrate it,but my ans turn out to be wrong..so could you please rectify my mistakes..thanks
If you have
\frac{4x^2+3x+6}{x^2(x+2)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+2}
then multiplying through gives
4x^2 + 3x + 6 = Ax(x+2) + B(x+2) + Cx^2.
Now you can equate coefficients as you intended to.
Unco the method which you did is the same which noobie has presented. It contains too many x's. There is an another nice way of doing it.
Try to obtain the numerator as a derivative of the denominator. And then the extra term which you get try to break it in simple parts.
ok,i understand..thanks a lot..
ok,i understand..thanks a lot..
Understood! What did you do with the remaining term?
vBulletin® v3.8.7, Copyright ©2000-2012, vBulletin Solutions, Inc.