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Integration using substitution

  1. Apr 10, 2015 #1
    1. The problem statement, all variables and given/known data
    $$\int\frac{x^2+3}{x^6(x^2+1)}dx$$

    2. Relevant equations

    None
    3. The attempt at a solution

    I easily got the answer using partial fractions by splitting the integral as ##\frac{Ax+B}{x^2+1}+\frac{C}{x}+\frac{D}{x^2}+\frac{E}{x^3}+...+\frac{H}{x^6}## and then finding the coefficients by equating the numerator ##x^2+3## with the one obtained using partial fractions by taking LCM.
    But is there an easier method?
     
  2. jcsd
  3. Apr 10, 2015 #2

    Mark44

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    The only "easier" method I can think of would be an ordinary substitution. Maybe there is one, but if there is, I don't see it. For this integral I would do exactly what you did.
     
  4. Apr 10, 2015 #3

    Zondrina

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    Break up the integral:

    $$\int \frac{x^2 + 3}{x^6(x^2 + 1)} dx = \int \frac{1}{x^4(x^2 + 1)} dx + \int \frac{3}{x^6(x^2 + 1)} dx$$
     
  5. Apr 10, 2015 #4

    SammyS

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    Maybe a different split:
    ##\displaystyle\ \frac{x^2+3}{x^6(x^2+1)}=\frac{x^2+1}{x^6(x^2+1)}+\frac{2}{x^6(x^2+1)}=\frac{1}{x^6}+\frac{2}{x^6(x^2+1)}\ ##

    Then noticing that the title of the thread is "Integration using substitution" and not "Integration by substitution" ... (maybe not intended but try it)

    Simplify ##\displaystyle\ \frac{2}{x^6(x^2+1)}\ ## by letting ##\ u=x^2\ .## Just do this to aid in manipulating the rational expression.

    This gives: ##\displaystyle\ \frac{2}{u^3(u+1)}\ ##

    Do partial fraction decomposition, or use the following reduction:
    ##\displaystyle\ \frac{1}{u^n(u+1)}=\frac{1}{u^n}-\frac{1}{u^{n-1}(u+1)}\ ## (repeatedly).

    Of course change ##\ u\ ## back to ##\ x^2\ ## before integrating.
     
  6. Apr 10, 2015 #5
    You can split it like this (better method): $$\frac{3x^2+3-2x^2}{x^6(x^2+1)}=\frac{3(x^2+1)-2x^2}{x^6(x^2+1)}=\frac{3(x^2+1)}{x^6(x^2+1)}-2\frac{x^2}{x^6(x^2+1)}=3\frac{1}{x^6}-2\frac{1}{x^4(x^2+1)}$$
     
  7. Apr 11, 2015 #6

    SammyS

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    I hesitate to put this up, but can't resist.
    ##\displaystyle\ \frac{3+x^2}{x^6(1+x^2)}=\frac{3+3x^2-2x^2}{x^6(1+x^2)} ##

    ##\displaystyle\ =\frac{3+3x^2-2x^2-2x^4+2x^4+2x^6-2x^6}{x^6(1+x^2)} ##

    ##\displaystyle\ =\frac{3(1+x^2)-2x^2(1+x^2)+2x^4(1+x^2)-2x^6}{x^6(1+x^2)} ##​

    Just a different way to get the partial fraction decomposition, I suppose.
     
  8. Apr 11, 2015 #7
    You get the the same thing if you use partial fractions like i did in post 0#
     
  9. Apr 11, 2015 #8

    SammyS

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    I would hope so.

    The partial fraction method often seems rather tedious to me. I'm not sure this was any better.

    Doing the substitution, u = x2 like I suggest in post #4 above could make the partial fraction thing easier, especially after using the split you suggested.
     
    Last edited: Apr 11, 2015
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