anonymous
  • anonymous
I'm having trouble with some partial fraction decomposition problems. I have to write out the form of the partial fraction of the function Q = limit(1 to 10) 1x/(x^2+6x+9) dx. Determine the numerival values of the coefficients, A and B ----A/denominator + B/denominator
Mathematics
schrodinger
  • schrodinger
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triciaal
  • triciaal
start by finding the factors of the denominator
triciaal
  • triciaal
|dw:1439752537963:dw|
anonymous
  • anonymous
I did that and got (x+3)^2

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anonymous
  • anonymous
thats the answer
anonymous
  • anonymous
Is it really?? I thought I had to reduce after that point
anonymous
  • anonymous
And what about the limit? The homework program I'm using says that's wrong
anonymous
  • anonymous
well it is wrong stop looking for handouts duhhh
triciaal
  • triciaal
do you need to integrate the function and find value of 10 minus value of 1?
anonymous
  • anonymous
I'm not, I just need help getting to the next point. I had factored it, then put it in the form \[\frac{ 1x }{ (x+3)(x+3) } = \frac{ A }{ (x+3) } + \frac{ B }{ (x+3) }\]
anonymous
  • anonymous
Then I guess I'm supposed to find where x = 0 and solve for A then insert to find B, but I'm having a tough time just isolating just one, normally the factors wouldn't be the same thing
anonymous
  • anonymous
I've just been struggling on this problem for a while now
triciaal
  • triciaal
triciaal
  • triciaal
how do you know the denominator for A or B is not (x +3)^2
triciaal
  • triciaal
|dw:1439753193971:dw|
anonymous
  • anonymous
|dw:1439753340124:dw|
anonymous
  • anonymous
Wow drawing is hard on this
ganeshie8
  • ganeshie8
Heyy
anonymous
  • anonymous
Hi
ganeshie8
  • ganeshie8
you need to square the second one, try this \[\frac{ 1x }{ (x+3)(x+3) } = \frac{ A }{ (x+3) } + \frac{ B }{ (x+3)^{\color{red}{2}} }\]
anonymous
  • anonymous
B = 1x - A?
misty1212
  • misty1212
then \[A(x+3)^2+B(x+3)=x\] is the next step you need the square when you have repeated factors in the denominator
misty1212
  • misty1212
oh wait i think i made a mistake there
misty1212
  • misty1212
when you add you should get \[A(x+3)+B\] as your numerator, so \[A(x+3)+B=x\]
misty1212
  • misty1212
multiply out, get \[Ax+3A+B=x\]telling you \(A=1\)
ganeshie8
  • ganeshie8
and you can get the value of \(B\) easily, simply plugin \(x=-3\)
misty1212
  • misty1212
and \[3A+B=0\]so \[3+B=0\\ B=-3\]
misty1212
  • misty1212
or @ganeshie8 way works too
anonymous
  • anonymous
Oh my goodness thank you guys so much!
misty1212
  • misty1212
\[\color\magenta\heartsuit\]
anonymous
  • anonymous
How do I close this problem? First question I've asked on here...
misty1212
  • misty1212
you should see an option in the bottom for "close question"

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