- anonymous

Partial fraction decompostion with integrals..

- katieb

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- anonymous

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

- anonymous

|dw:1360273420317:dw|

- anonymous

stuck here

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## More answers

- anonymous

i need to find A B and C

- anonymous

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?

- anonymous

|dw:1360273719414:dw|

- anonymous

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

- anonymous

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

- anonymous

|dw:1360274047172:dw|

- anonymous

Yep

- anonymous

|dw:1360274149320:dw|

- anonymous

so far so good?

- anonymous

mhmm

- anonymous

|dw:1360274464556:dw|

- anonymous

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

- anonymous

oh ya!

- anonymous

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

- anonymous

my online homework still says im wrong

- anonymous

maybe i need to combine the ln's?

- anonymous

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?

- anonymous

the x^2

- anonymous

a/x + b/x^2

- anonymous

im on my last attempt to answer it correctly , damn

- anonymous

@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.

- anonymous

@SithsAndGiggles can you see where I went wrong?

- anonymous

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.

- anonymous

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
\[-9=Ax(x-3)+B(x-3)+Cx^2\\
-9=(A+C)x^2+(-3A+B)x-3B\\
\text{Matching up coefficients, you get}\\
\begin{cases}A+C=0\\
-3A+B=0\\
-3B=-9\end{cases}\\
\text{Solving, you get}\\
B=3, A=1, C=-1\]

- anonymous

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

- anonymous

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.

- anonymous

thanks a ton!

- anonymous

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

- anonymous

NOOOOOOOOOOOO

- anonymous

thank you so much for the help!!!

- anonymous

cant believe i missed that!!!

- anonymous

|dw:1360279683216:dw|

- anonymous

err ln2 instead of ln3

- anonymous

and no + C

- anonymous

its right!!!!!!!!!

- anonymous

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

- anonymous

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

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