## dfresenius 2 years ago Partial fraction decompostion with integrals..

1. dfresenius

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

2. dfresenius

|dw:1360273420317:dw|

3. dfresenius

stuck here

4. dfresenius

i need to find A B and C

5. Jemurray3

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

|dw:1360273719414:dw|

7. dfresenius

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

8. Jemurray3

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

9. dfresenius

|dw:1360274047172:dw|

10. Jemurray3

Yep

11. dfresenius

|dw:1360274149320:dw|

12. dfresenius

so far so good?

13. Jemurray3

mhmm

14. dfresenius

|dw:1360274464556:dw|

15. Jemurray3

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

16. dfresenius

oh ya!

17. Jemurray3

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

18. dfresenius

my online homework still says im wrong

19. dfresenius

maybe i need to combine the ln's?

20. cherio12

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

the x^2

22. dfresenius

a/x + b/x^2

23. dfresenius

im on my last attempt to answer it correctly , damn

24. SithsAndGiggles

@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

@SithsAndGiggles can you see where I went wrong?

26. SithsAndGiggles

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

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

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

29. SithsAndGiggles

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

thanks a ton!

31. SithsAndGiggles

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

32. dfresenius

NOOOOOOOOOOOO

33. dfresenius

thank you so much for the help!!!

34. dfresenius

cant believe i missed that!!!

35. dfresenius

|dw:1360279683216:dw|

36. dfresenius

37. dfresenius

and no + C

38. dfresenius

its right!!!!!!!!!

39. SithsAndGiggles

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

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