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anonymous
 3 years ago
did i get this partial integration question right?
anonymous
 3 years ago
did i get this partial integration question right?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1359940501331:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Answer dw:1359940559335:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can someone just check if that is the right answer

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0im still kind of confused about it

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0i think you've made a mistake somewhere,

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok i did it wrong apparently but i dont know how to do it still :( ?

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1so, denominator = x^2 (x4) do you know partial fractions method ? also, if you show your work, we can quickly spot the error, and continue from that step...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok i will type it out for you

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so first what i did was split the denominator to x(x^24) then split that to (x+2)(x2)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then set the 2x^2 12x+4=A/x+B/(x+2)+C/(x2)

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1here.... x^34x^2 = x^2 (x4) and not x (x^24)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wow i cant believe i missed that

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1since you know partial fractions , i think you'll get it in your next try..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sorry guys my intentet is acting weird

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yea ill try the steps again and after im done pm you the steps

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0to see if i did it right thanks

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1ok, sure. or you can post it here.. welcome ^_^

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so would the partial thing be Ax+ B /x^2 +C/(x4)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and really sorrmy internet is going berserk for some reason today college inter is acting funky for some reason

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wait isnt when the denominator is x^2 the top part is (Ax + B )/(x^2)

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1its the same thing! (Ax+B)/x^2 = A/x + B/x^2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yep really had a long day today . after the superbowl drink and math dont go together

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok after im done ill post the answer ty

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so 2x^212x+4 = A/x + B/x^2 +c/(x4)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and i solved for c which got me 3/16

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1359956605483:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and i plugged in as x = 4;

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0to get c = 3/16 but you cant solve for the other A and B is there another method to do this? i dont get partial fractions

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1to find c, put x=4 to find b, put x=0 to find a, put x=any other value, say x=1

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1what part of partial fractions, do you not get ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if you put x = 0 you will get all 3 = 0

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1359957041709:dw so you cant put 0

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yea i got the c one though with is at x = 4

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1A x (x4) +B (x4) +C (x^2)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1359957163440:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wait can you tell me how you got that i got the trigonometric integral stuff but this stuff is confusing me

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so can you tell me how you cancelled out the x in the B and the x in the A

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0shouldnt A/x be multiplied by dw:1359957703525:dw

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1then denominator of A will become x^3 (x4) which we don't want and is not common

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hold on what i meant was isnt the format dw:1359957835056:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1359958002008:dw i guess what i mean is in this for example why did you only pick out x(x4) for A only (x4) for b and x^2 for C im really confused about that

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1so, that i have SAME denominator for all the 3.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and also because in previous problems i did it was like 4x^2 + 54 x + 134 = A/(x1) +B/(x+5) +C/(x+3) which i got to 4x^2 + 54x + 134 = A(x+5)(x+3) + B(x1)(x+3)+C(x1)(x+5)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0do you have to do this when the denominator has a square on the x? or is it dependent on the problem

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1359958265843:dwits as simple as this...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh i see i guess you could multiply what i did but it would cancel

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i think i finally get it now

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1in previous problem, to make the common denominator as (x5)(x+5)(x+3) so, you multiplied and divided by (x+5)(x+3) for A yes, you can say that also...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now if i had payed attention in my algebra classes this would be so much easier .

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1no problem! because its easier for you now! right ?
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