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

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dfresenius
 2 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{4}^{5}\frac{ x^33x^2 9}{ x^33x^2 }dx\]

dfresenius
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1360273420317:dw

dfresenius
 2 years ago
Best ResponseYou've already chosen the best response.0i need to find A B and C

Jemurray3
 2 years ago
Best ResponseYou've already chosen the best response.1Ignore the one for now, just try to decompose \[ \frac{9}{x^2(x3)} =\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x3} \] Multiply through by x^2(x3) to get \[ 9 = Ax(x3)+ B(x3) + Cx^2 \] There is a tricky way, which is faster, or a straightforward way, which takes longer. Which would you prefer?

dfresenius
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1360273719414:dw

dfresenius
 2 years ago
Best ResponseYou've already chosen the best response.0so, now, I can plug those in to find A?

Jemurray3
 2 years ago
Best ResponseYou've already chosen the best response.1Plug those values in for B and C and just pick any arbitrary value of x to get A.

dfresenius
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1360274047172:dw

dfresenius
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1360274149320:dw

dfresenius
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1360274464556:dw

Jemurray3
 2 years ago
Best ResponseYou've already chosen the best response.1Shouldn't the 3rd and 6th terms be ln(2) and ln(1)?

Jemurray3
 2 years ago
Best ResponseYou've already chosen the best response.1Other than that, looks fine. you can write ln(4) as 2 ln(2) if you'd like.

dfresenius
 2 years ago
Best ResponseYou've already chosen the best response.0my online homework still says im wrong

dfresenius
 2 years ago
Best ResponseYou've already chosen the best response.0maybe i need to combine the ln's?

cherio12
 2 years ago
Best ResponseYou've already chosen the best response.0I think you decompose it to\[\frac{ A }{ x3 }+\frac{ B+Cx }{ x^2 }\]..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?

dfresenius
 2 years ago
Best ResponseYou've already chosen the best response.0im on my last attempt to answer it correctly , damn

SithsAndGiggles
 2 years ago
Best ResponseYou've already chosen the best response.0@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.

dfresenius
 2 years ago
Best ResponseYou've already chosen the best response.0@SithsAndGiggles can you see where I went wrong?

SithsAndGiggles
 2 years ago
Best ResponseYou've already chosen the best response.0Your 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.

SithsAndGiggles
 2 years ago
Best ResponseYou've already chosen the best response.0Okay, so using the longer method mentioned by @Jemurray3, I get different values of A, B, and C: \[1 + \int_{4}^{5}\frac{9}{x^2(x3)}dx\] Focusing on the integral: Decomposing into partial fractions, you get \[9=Ax(x3)+B(x3)+Cx^2\\ 9=(A+C)x^2+(3A+B)x3B\\ \text{Matching up coefficients, you get}\\ \begin{cases}A+C=0\\ 3A+B=0\\ 3B=9\end{cases}\\ \text{Solving, you get}\\ B=3, A=1, C=1\]

dfresenius
 2 years ago
Best ResponseYou've already chosen the best response.0strange, how will I know when to use each method?

SithsAndGiggles
 2 years ago
Best ResponseYou've already chosen the best response.0The 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.

SithsAndGiggles
 2 years ago
Best ResponseYou've already chosen the best response.0Yep, that's what it was.dw:1360279359173:dw

dfresenius
 2 years ago
Best ResponseYou've already chosen the best response.0thank you so much for the help!!!

dfresenius
 2 years ago
Best ResponseYou've already chosen the best response.0cant believe i missed that!!!

dfresenius
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1360279683216:dw

dfresenius
 2 years ago
Best ResponseYou've already chosen the best response.0err ln2 instead of ln3

SithsAndGiggles
 2 years ago
Best ResponseYou've already chosen the best response.0I'm not getting the ln4 that you are. Here's my work, postintegration: \[1+(\ln 5\ln 4)  3\left(\frac{1}{5}\frac{1}{4}\right)(\ln 2  \ln 1)\\ 1+\ln 52\ln 2 +\frac{3}{20}\ln 2\\ \ln 53\ln 2 +\frac{23}{20}\\ \ln 5\ln 8 +\frac{23}{20}\\ \ln {\frac{5}{8}}+\frac{23}{20}\\\]

SithsAndGiggles
 2 years ago
Best ResponseYou've already chosen the best response.0I take it back, it was there, I just rewrote it. You're welcome!
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