- 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

- schrodinger

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- triciaal

start by finding the factors of the denominator

- triciaal

|dw:1439752537963:dw|

- anonymous

I did that and got (x+3)^2

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## More answers

- anonymous

thats the answer

- anonymous

Is it really?? I thought I had to reduce after that point

- anonymous

And what about the limit? The homework program I'm using says that's wrong

- anonymous

well it is wrong stop looking for handouts duhhh

- triciaal

do you need to integrate the function and find value of 10 minus value of 1?

- 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

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

I've just been struggling on this problem for a while now

- triciaal

- triciaal

how do you know the denominator for A or B is not (x +3)^2

- triciaal

|dw:1439753193971:dw|

- anonymous

|dw:1439753340124:dw|

- anonymous

Wow drawing is hard on this

- ganeshie8

Heyy

- anonymous

Hi

- 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

B = 1x - A?

- 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

oh wait i think i made a mistake there

- misty1212

when you add you should get
\[A(x+3)+B\] as your numerator, so
\[A(x+3)+B=x\]

- misty1212

multiply out, get \[Ax+3A+B=x\]telling you \(A=1\)

- ganeshie8

and you can get the value of \(B\) easily, simply plugin \(x=-3\)

- misty1212

and \[3A+B=0\]so
\[3+B=0\\
B=-3\]

- misty1212

or @ganeshie8 way works too

- anonymous

Oh my goodness thank you guys so much!

- misty1212

\[\color\magenta\heartsuit\]

- anonymous

How do I close this problem? First question I've asked on here...

- misty1212

you should see an option in the bottom for "close question"

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