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Partial fraction decompostion with integrals..

Mathematics
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\[\int\limits_{4}^{5}\frac{ x^3-3x^2 -9}{ x^3-3x^2 }dx\]
|dw:1360273420317:dw|
stuck here

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Other answers:

i need to find A B and C
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?
|dw:1360273719414:dw|
so, now, I can plug those in to find A?
Plug those values in for B and C and just pick any arbitrary value of x to get A.
|dw:1360274047172:dw|
Yep
|dw:1360274149320:dw|
so far so good?
mhmm
|dw:1360274464556:dw|
Shouldn't the 3rd and 6th terms be ln(2) and ln(1)?
oh ya!
Other than that, looks fine. you can write ln(4) as 2 ln(2) if you'd like.
my online homework still says im wrong
maybe i need to combine the ln's?
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?
the x^2
a/x + b/x^2
im on my last attempt to answer it correctly , damn
@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.
@SithsAndGiggles can you see where I went wrong?
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.
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 \[-9=Ax(x-3)+B(x-3)+Cx^2\\ -9=(A+C)x^2+(-3A+B)x-3B\\ \text{Matching up coefficients, you get}\\ \begin{cases}A+C=0\\ -3A+B=0\\ -3B=-9\end{cases}\\ \text{Solving, you get}\\ B=3, A=1, C=-1\]
strange, how will I know when to use each method?
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.
thanks a ton!
Yep, that's what it was.|dw:1360279359173:dw|
NOOOOOOOOOOOO
thank you so much for the help!!!
cant believe i missed that!!!
|dw:1360279683216:dw|
err ln2 instead of ln3
and no + C
its right!!!!!!!!!
I'm not getting the -ln4 that you are. Here's my work, post-integration: \[1+(\ln 5-\ln 4) - 3\left(\frac{1}{5}-\frac{1}{4}\right)-(\ln 2 - \ln 1)\\ 1+\ln 5-2\ln 2 +\frac{3}{20}-\ln 2\\ \ln 5-3\ln 2 +\frac{23}{20}\\ \ln 5-\ln 8 +\frac{23}{20}\\ \ln {\frac{5}{8}}+\frac{23}{20}\\\]
I take it back, it was there, I just rewrote it. You're welcome!

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