## anonymous one year ago 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

1. triciaal

start by finding the factors of the denominator

2. triciaal

|dw:1439752537963:dw|

3. anonymous

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

4. anonymous

5. anonymous

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

6. anonymous

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

7. anonymous

well it is wrong stop looking for handouts duhhh

8. triciaal

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

9. anonymous

I'm not, I just need help getting to the next point. I had factored it, then put it in the form $\frac{ 1x }{ (x+3)(x+3) } = \frac{ A }{ (x+3) } + \frac{ B }{ (x+3) }$

10. 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

11. anonymous

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

12. triciaal

@zepdrix

13. triciaal

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

14. triciaal

|dw:1439753193971:dw|

15. anonymous

|dw:1439753340124:dw|

16. anonymous

Wow drawing is hard on this

17. ganeshie8

Heyy

18. anonymous

Hi

19. ganeshie8

you need to square the second one, try this $\frac{ 1x }{ (x+3)(x+3) } = \frac{ A }{ (x+3) } + \frac{ B }{ (x+3)^{\color{red}{2}} }$

20. anonymous

B = 1x - A?

21. 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

22. misty1212

oh wait i think i made a mistake there

23. misty1212

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

24. misty1212

multiply out, get $Ax+3A+B=x$telling you $$A=1$$

25. ganeshie8

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

26. misty1212

and $3A+B=0$so $3+B=0\\ B=-3$

27. misty1212

or @ganeshie8 way works too

28. anonymous

Oh my goodness thank you guys so much!

29. misty1212

$\color\magenta\heartsuit$

30. anonymous

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

31. misty1212

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