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noobie!
Jan1-09, 02:32 AM
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:

FedEx
Jan1-09, 09:50 AM
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.

Unco
Jan1-09, 02:09 PM
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.

FedEx
Jan2-09, 03:51 AM
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.

noobie!
Jan3-09, 10:36 AM
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..

FedEx
Jan3-09, 10:44 AM
ok,i understand..thanks a lot..

Understood! What did you do with the remaining term?