dfresenius Group Title Partial fraction decompostion with integrals.. one year ago one year ago

1. dfresenius Group Title

$\int\limits_{4}^{5}\frac{ x^3-3x^2 -9}{ x^3-3x^2 }dx$

2. dfresenius Group Title

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3. dfresenius Group Title

stuck here

4. dfresenius Group Title

i need to find A B and C

5. Jemurray3 Group Title

Ignore the one for now, just try to decompose $\frac{9}{x^2(x-3)} =\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-3}$ Multiply through by x^2(x-3) to get $9 = Ax(x-3)+ B(x-3) + Cx^2$ There is a tricky way, which is faster, or a straightforward way, which takes longer. Which would you prefer?

6. dfresenius Group Title

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7. dfresenius Group Title

so, now, I can plug those in to find A?

8. Jemurray3 Group Title

Plug those values in for B and C and just pick any arbitrary value of x to get A.

9. dfresenius Group Title

|dw:1360274047172:dw|

10. Jemurray3 Group Title

Yep

11. dfresenius Group Title

|dw:1360274149320:dw|

12. dfresenius Group Title

so far so good?

13. Jemurray3 Group Title

mhmm

14. dfresenius Group Title

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15. Jemurray3 Group Title

Shouldn't the 3rd and 6th terms be ln(2) and ln(1)?

16. dfresenius Group Title

oh ya!

17. Jemurray3 Group Title

Other than that, looks fine. you can write ln(4) as 2 ln(2) if you'd like.

18. dfresenius Group Title

my online homework still says im wrong

19. dfresenius Group Title

maybe i need to combine the ln's?

20. cherio12 Group Title

I think you decompose it to$\frac{ A }{ x-3 }+\frac{ B+Cx }{ x^2 }$..not a 100% sure..but there is something different done when you are over a power greater than one. Side note, were did you get A/x from?

21. dfresenius Group Title

the x^2

22. dfresenius Group Title

a/x + b/x^2

23. dfresenius Group Title

im on my last attempt to answer it correctly , damn

24. SithsAndGiggles Group Title

@cherio12, when decomposing (something)/x^2 you have $\frac{A}{x}+\frac{B}{x^2}$ I'm not familiar with the formal explanation as to why this happens, but I know if you have factors with multiplicity greater than 1 they're treated as normal linear factors.

25. dfresenius Group Title

@SithsAndGiggles can you see where I went wrong?

26. SithsAndGiggles Group Title

Your answer doesn't look wrong at first glance, but I'm still working back. It could be that the program your class uses is very picky about the answer it accepts.

27. SithsAndGiggles Group Title

Okay, so using the longer method mentioned by @Jemurray3, I get different values of A, B, and C: $1 + \int_{4}^{5}-\frac{9}{x^2(x-3)}dx$ Focusing on the integral: Decomposing into partial fractions, you get $-9=Ax(x-3)+B(x-3)+Cx^2\\ -9=(A+C)x^2+(-3A+B)x-3B\\ \text{Matching up coefficients, you get}\\ \begin{cases}A+C=0\\ -3A+B=0\\ -3B=-9\end{cases}\\ \text{Solving, you get}\\ B=3, A=1, C=-1$

28. dfresenius Group Title

strange, how will I know when to use each method?

29. SithsAndGiggles Group Title

The method I used works pretty much every time. You're less prone to making mistakes, I think. I'll check your work again; if the answer really is wrong, it'll probably be an algebra mistake.

30. dfresenius Group Title

thanks a ton!

31. SithsAndGiggles Group Title

Yep, that's what it was.|dw:1360279359173:dw|

32. dfresenius Group Title

NOOOOOOOOOOOO

33. dfresenius Group Title

thank you so much for the help!!!

34. dfresenius Group Title

cant believe i missed that!!!

35. dfresenius Group Title

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36. dfresenius Group Title

37. dfresenius Group Title

and no + C

38. dfresenius Group Title

its right!!!!!!!!!

39. SithsAndGiggles Group Title

I'm not getting the -ln4 that you are. Here's my work, post-integration: $1+(\ln 5-\ln 4) - 3\left(\frac{1}{5}-\frac{1}{4}\right)-(\ln 2 - \ln 1)\\ 1+\ln 5-2\ln 2 +\frac{3}{20}-\ln 2\\ \ln 5-3\ln 2 +\frac{23}{20}\\ \ln 5-\ln 8 +\frac{23}{20}\\ \ln {\frac{5}{8}}+\frac{23}{20}\\$

40. SithsAndGiggles Group Title

I take it back, it was there, I just rewrote it. You're welcome!