anonymous
  • anonymous
did i get this partial integration question right?
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
|dw:1359940501331:dw|
anonymous
  • anonymous
Answer |dw:1359940559335:dw|
anonymous
  • anonymous
can someone just check if that is the right answer

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anonymous
  • anonymous
im still kind of confused about it
UnkleRhaukus
  • UnkleRhaukus
i think you've made a mistake somewhere,
anonymous
  • anonymous
ok i did it wrong apparently but i dont know how to do it still :( ?
hartnn
  • hartnn
so, denominator = x^2 (x-4) do you know partial fractions method ? also, if you show your work, we can quickly spot the error, and continue from that step...
anonymous
  • anonymous
ok i will type it out for you
anonymous
  • anonymous
so first what i did was split the denominator to x(x^2-4) then split that to (x+2)(x-2)
anonymous
  • anonymous
then set the 2x^2 -12x+4=A/x+B/(x+2)+C/(x-2)
hartnn
  • hartnn
here.... x^3-4x^2 = x^2 (x-4) and not x (x^2-4)
anonymous
  • anonymous
wow i cant believe i missed that
hartnn
  • hartnn
since you know partial fractions , i think you'll get it in your next try..
anonymous
  • anonymous
sorry guys my intentet is acting weird
anonymous
  • anonymous
yea ill try the steps again and after im done pm you the steps
anonymous
  • anonymous
to see if i did it right thanks
hartnn
  • hartnn
ok, sure. or you can post it here.. welcome ^_^
anonymous
  • anonymous
so would the partial thing be Ax+ B /x^2 +C/(x-4)
hartnn
  • hartnn
yes.
anonymous
  • anonymous
and really sorrmy internet is going berserk for some reason today college inter is acting funky for some reason
hartnn
  • hartnn
A/x+ B /x^2 +C/(x-4)
anonymous
  • anonymous
wait isnt when the denominator is x^2 the top part is (Ax + B )/(x^2)
hartnn
  • hartnn
its the same thing! (Ax+B)/x^2 = A/x + B/x^2
anonymous
  • anonymous
yep really had a long day today -.- after the superbowl drink and math dont go together
hartnn
  • hartnn
lol.
anonymous
  • anonymous
ok after im done ill post the answer ty
anonymous
  • anonymous
so 2x^2-12x+4 = A/x + B/x^2 +c/(x-4)
anonymous
  • anonymous
and i solved for c which got me -3/16
anonymous
  • anonymous
|dw:1359956605483:dw|
anonymous
  • anonymous
and i plugged in as x = 4;
anonymous
  • anonymous
to 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
  • hartnn
to find c, put x=4 to find b, put x=0 to find a, put x=any other value, say x=1
hartnn
  • hartnn
what part of partial fractions, do you not get ?
anonymous
  • anonymous
if you put x = 0 you will get all 3 = 0
hartnn
  • hartnn
u sure ?
anonymous
  • anonymous
|dw:1359957041709:dw| so you cant put 0
anonymous
  • anonymous
yea i got the c one though with is at x = 4
hartnn
  • hartnn
......+B (x-4)+.....
hartnn
  • hartnn
A x (x-4) +B (x-4) +C (x^2)
anonymous
  • anonymous
|dw:1359957163440:dw|
anonymous
  • anonymous
wait can you tell me how you got that i got the trigonometric integral stuff but this stuff is confusing me
anonymous
  • anonymous
so can you tell me how you cancelled out the x in the B and the x in the A
hartnn
  • hartnn
|dw:1359957466093:dw|
hartnn
  • hartnn
|dw:1359957564998:dw|
anonymous
  • anonymous
shouldnt A/x be multiplied by |dw:1359957703525:dw|
hartnn
  • hartnn
then denominator of A will become x^3 (x-4) which we don't want and is not common
anonymous
  • anonymous
hold on what i meant was isnt the format |dw:1359957835056:dw|
anonymous
  • anonymous
|dw:1359958002008:dw| i guess what i mean is in this for example why did you only pick out x(x-4) for A only (x-4) for b and x^2 for C im really confused about that
hartnn
  • hartnn
so, that i have SAME denominator for all the 3.
anonymous
  • anonymous
and also because in previous problems i did it was like 4x^2 + 54 x + 134 = A/(x-1) +B/(x+5) +C/(x+3) which i got to 4x^2 + 54x + 134 = A(x+5)(x+3) + B(x-1)(x+3)+C(x-1)(x+5)
anonymous
  • anonymous
do you have to do this when the denominator has a square on the x? or is it dependent on the problem
hartnn
  • hartnn
|dw:1359958265843:dw|its as simple as this...
hartnn
  • hartnn
do what ?
anonymous
  • anonymous
oh i see i guess you could multiply what i did but it would cancel
anonymous
  • anonymous
i think i finally get it now
hartnn
  • hartnn
in previous problem, to make the common denominator as (x-5)(x+5)(x+3) so, you multiplied and divided by (x+5)(x+3) for A yes, you can say that also...
anonymous
  • anonymous
now if i had payed attention in my algebra classes this would be so much easier -.-
hartnn
  • hartnn
no problem! because its easier for you now! right ?

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