Kkutie7
  • Kkutie7
Please make sure I'm doing this right. \[\int\frac{3y^{3}+5y-1}{y^{3}+y}dy\]
Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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Kkutie7
  • Kkutie7
\[\rightarrow \int\frac{3y^{3}+5y-1}{y(y^{2}-1)}dy\]
Kkutie7
  • Kkutie7
\[\rightarrow\int\frac{3y^{3}+5y-1}{y(y-1)(y+1)}dy\] \[\rightarrow \frac{3y^{3}+5y-1}{y(y-1)(y+1)}=\frac{A}{y}+\frac{B}{y-1}+\frac{C}{y+1}\]
Kkutie7
  • Kkutie7
\[\rightarrow 3y^{3}+5y-1=A(y-1)(y+1)+B(y)(y+1)+C(y)(y-1)\]

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anonymous
  • anonymous
Using partial fraction decomposition for this program will be messy. I haven't worked out all the details but here's what I have so far... I'm pretty sure all of my steps are legal... just breaking up the numerator more and more until we get rid of y^3 altogether.
Kkutie7
  • Kkutie7
\[\rightarrow x=0\rightarrow -1=-A, A=1\] \[\rightarrow x=1,3+5-1=2B, B=\frac{7}{2}\] \[\rightarrow x=-1,2C, C=\frac{-1}{2} \] I don't see how things are messy, unless I'm doing it wrong
Kkutie7
  • Kkutie7
\[\rightarrow \int\frac{1}{y}+\frac{7}{2}\int\frac{1}{y-1}+\frac{-1}{2}\int\frac{1}{y+1}\] \
Kkutie7
  • Kkutie7
\[\rightarrow ln|y|+\frac{7}{2}ln|y-1|-\frac{1}{2}ln|y+1|+C\] is this right?
jim_thompson5910
  • jim_thompson5910
A = 1 ... agree B = 7/2 ... agree C = -1/2 ... disagree
Kkutie7
  • Kkutie7
C=-9/2?
jim_thompson5910
  • jim_thompson5910
nope
Kkutie7
  • Kkutie7
damn ok hold on
ganeshie8
  • ganeshie8
It is all wrong. You must do long division first and make the degree of numerator less than the degree of denominator..
jim_thompson5910
  • jim_thompson5910
@ganeshie8 no it's not all wrong. The only mistake I see is that C is the only thing that's off. Everything else looks fine
Kkutie7
  • Kkutie7
so:|dw:1449894862370:dw|
Kkutie7
  • Kkutie7
|dw:1449895145767:dw|
jim_thompson5910
  • jim_thompson5910
oh my mistake. I thought the 3y^3 was 3y^2 for some reason
mathmale
  • mathmale
there's more than 1 way in which to skin a cat. You've used partial fractions: you've ended up with a partial fraction expansion of the original integrand. Very good. But you could also use ganeshi8's suggestion: You must do long division first and make the degree of numerator less than the degree of denominator.. some of the steps you'd go through in this second method would be similar to those of your complete partial fraction expansion.
Kkutie7
  • Kkutie7
well i just looked at my answer key and it said i must use long division first so im getting a little confused
jim_thompson5910
  • jim_thompson5910
you're correct @Kkutie7 from this page http://tutorial.math.lamar.edu/Classes/Alg/PartialFractions.aspx it says `Partial fractions can only be done if the degree of the numerator is strictly less than the degree of the denominator` In this case, both degrees are 3 so long division must be done first then do partial fraction decomposition on the remainder portion
Kainui
  • Kainui
By the way, you might find www.wolframalpha.com to be a useful resource @Kkutie7 it's pretty powerful. Like say your answer doesn't match the answer in the book or whatever, you can type it in like this to check to see if it's equivalent: http://www.wolframalpha.com/input/?i=x%2F2%2Bx%5E2%2F2%20%3D%20x(x%2B1)%2F2&t=crmtb01 Just some sample input: http://www.wolframalpha.com/input/?i=integral+1%2Fsqrt%7Bx%5E2-12%7D
Kkutie7
  • Kkutie7
oh yeah I use it all the time. thanks

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