Luigi0210
  • Luigi0210
I need a bit of help with integration
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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Luigi0210
  • Luigi0210
\[\int\limits \frac{ 4x^3+3x^2+2x+1 }{ x^4-1 }\]
Luigi0210
  • Luigi0210
*dx
anonymous
  • anonymous
Looks a lot like you want to do some partial fraction decomposition at first, maybe a long hand division at first :-)

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zepdrix
  • zepdrix
\[\int\limits\frac{4x^3+3x^2+2x+1}{x^4-1} \qquad = \qquad \int\limits \frac{4x^3}{x^4-1}dx+\int\limits \frac{3x^2+2x+1}{x^4-1}dx\] Hmm If we split up the fraction like this, I think we can do a nice easy U-sub on the first integral. Then for the other one, maybe some factoring, hmm.
zepdrix
  • zepdrix
Err ya I guess partial fractions would work better :\
Luigi0210
  • Luigi0210
Yea, so how would I start this off?
zepdrix
  • zepdrix
Start by factoring the bottom: We have the difference of squares, \[\large x^4-1 \qquad = \qquad (x^2)^2-(1)^2\]Remember how to break down the difference of squares into factors?
Luigi0210
  • Luigi0210
(x^2-1)(x^2+1)?
zepdrix
  • zepdrix
Good good, looks like we can break down that first set of brackets by repeating the rule :)
Luigi0210
  • Luigi0210
\[\int\limits \frac{ 4x^3+3x^2+2x+1 }{ (x+1)(x-1)(x^2+1) } dx\]
zepdrix
  • zepdrix
looks good. do you understand how to do the initial setup for the partial fractions?
Luigi0210
  • Luigi0210
\[\frac{ 4x^3+3x^2+2x+1 }{ (x-1)(x+1)(x^2+1) }=\frac{ A }{ (x-1) }+\frac{ B }{ (x+1) }+\frac{ Cx+D }{ x^2+1 }\]
Loser66
  • Loser66
go ahead, friend. you are good
zepdrix
  • zepdrix
Yah good job :) looks correct so far.
Loser66
  • Loser66
@zepdrix If he steps up on the right track, give him medal, OK? hihi..
Luigi0210
  • Luigi0210
\[4x^3+3x^2+2x+1=A(x+1)(x^2+1)+B(x-1)(x^2+1)+(Cx+D)(x^2-1)\]
zepdrix
  • zepdrix
You might not want to combine those terms on the (Cx+D). The square will make it harder to solve for your constants. Well unless you plan on multiplying everything out, then I guess it doesn't matter. Depends what method you use.
Luigi0210
  • Luigi0210
that method confuses me
zepdrix
  • zepdrix
So if you leave the last term as (Cx+D)(x-1)(x+1), we can solve for A and B fairly easily, Let \(\large x=1\), and you can solve for A. Let \(\large x=-1\), and you can solve for B.
zepdrix
  • zepdrix
We might still have to multiply it all out to solve for C and D though.. hmm
Luigi0210
  • Luigi0210
So 10=4A A=2.5 ---------------------- -2=-4B B=.5
zepdrix
  • zepdrix
5/2 and 1/2? Ya, cool c:
zepdrix
  • zepdrix
We could probably solve for D by plugging on \(\large x=0\)
Luigi0210
  • Luigi0210
I got 1=C and D=1
zepdrix
  • zepdrix
cool, sounds correct.
Luigi0210
  • Luigi0210
So my only real problem is plugging all the info back into the integration equation..
zepdrix
  • zepdrix
Ok, it's not as bad as it seems. We want to plug everything back into this, \[\large \frac{A}{x-1}+\frac{B}{x+1}+\frac{Cx+D}{x^2+1}\] And then we'll take the integral of that, term by term.
Luigi0210
  • Luigi0210
So just plug in the values to the variables? Sorry about that delay

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