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[itex]\int\frac{n}{(n^{2}+1)^{2}}[/itex]= itself w/ Partial Fractions

  1. Nov 8, 2013 #1
    1. The problem statement, all variables and given/known data

    Why when I try to evaluate this with Partial Fractions, why do I end up with the original function?



    [itex]\frac{Ax+B}{n^{2}+1} + \frac{Cx+D}{(n^{2}+1)^{2}}[/itex]

    1n = (An+B)(n^{2}+1) + Cx + D

    [itex]0n^{3}+ 0n^{2} + 1n + 0n^{0} = n^{3}(A) + n^{2}(B) + n(A+C) + n^{0}(B+D)[/itex]

    A=0 B=0 C=1 D=0

    [itex]\frac{0n+0}{n^{2}+1} + \frac{1n+0}{(n^{2}+1)^{2}}[/itex]

    = 0 + [itex]\frac{n}{(n^{2}+1)^{2}}[/itex]
  2. jcsd
  3. Nov 8, 2013 #2


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  4. Nov 8, 2013 #3
    That website looks all scrambled up. Nothing seems to be aligned. Which part pertains to why this particular function ends up to simply be itself?

    By the way, when I do u-sub on this function I get 1/4 and when I do partial fractions I get -1/4. Is that a sign mistake?

    Also, the basis for this question was actually in regards to the series of the function in question.

    I was trying to do partial fractions to create a telescoping function to evaluate the series. But partial fractions doesn't separate this function into telescoping series.

    So my follow up questions is: Why does this function not separate into a telescoping form if Partial fractions can be applied to it?
  5. Nov 8, 2013 #4


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    What you're saying doesn't make a lot of sense. As you've seen, partial fractions leads you right back to the exact same integrand, so I don't see how you were able to carry out the integration that way. Substitution is the way to go here.

    The integral looks like it would be amenable to a trig substitution, but that seems like a lot of work when an ordinary substitution would do the trick.
  6. Nov 8, 2013 #5


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    The site works fine with Chrome on my laptop. Check your browser!

    The web site page goes over the method in detail, with a few examples. A review of the method may help you to find any problems with your technique.

    Wolfram Alpha says: ## n^2/(n^2+1) = 1-1/(n^2+1) ##
  7. Nov 9, 2013 #6


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    IDK why you want to use partial fractions anyway. You've plainly got a case for a u substitution, no partial fractions or trig substitution required.
  8. Nov 9, 2013 #7
    I don't see the need of a trig substitution. There is a very obvious substitution you can use as SteamKing has said.
  9. Nov 9, 2013 #8
    I just [thought I] saw an opportunity to practice a partial fraction from the past chapter while working on series.

    Thanks all.
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