Results 1 to 6 of 6

Math Help - quick integral conversion question

  1. #1
    Junior Member
    Joined
    Sep 2007
    Posts
    39

    quick integral conversion question

    Im looking at the solution to one of my teachers answers and just wondering how he got this:

    (X^4) / ({X^2}+1)dx is the same statement as X^2-1+[1/({X^2}+1)]dx

    Both are integrals obviously and i dont know how he got the 2nd from the 1st. Some sort of algebra/identity i assume? Thanks

    Chris
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Jun 2006
    From
    San Diego
    Posts
    101
    Quote Originally Posted by chrsr345 View Post
    Im looking at the solution to one of my teachers answers and just wondering how he got this:

    (X^4) / ({X^2}+1)dx is the same statement as X^2-1+[1/({X^2}+1)]dx

    Both are integrals obviously and i dont know how he got the 2nd from the 1st. Some sort of algebra/identity i assume? Thanks

    Chris
    Put the three terms in second expression under a common denominator and you'll get the first.

    The process of going form the first to the second is a bit more difficult, but it's called breaking a fraction into partial fractions.
    You basically do it by establishing an equation like:

    \frac{x^4}{x^2+1} = A + \frac{B}{x^2+1},

    then multiply by the denominator to get:

    x^4 = A(x^2+1) + B

    Expand:

    x^4 = Ax^2+(A + B)

    and compare terms on each side.
    Get A = x^2 and A+B=0, so B = -x^2.
    Therefore
    \frac{x^4}{x^2+1} = x^2- \frac{x^2}{x^2+1}

    This process can be repeated on the remaining fraction:

    \frac{x^2}{x^2+1} = C + \frac{D}{x^2 + 1}.
    Multiply to get:

    x^2 = C(x^2+1)+D = Cx^2 + (C+D).

    Compare terms again, and this time get C = 1, and D = -1.

    Put it all together:

    \frac{x^4}{x^2+1} = x^2- 1 + \frac{1}{x^2+1}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    14
    You don't even need partial fractions, you just need add & remove:

    \frac{{x^4 }}<br />
{{x^2  + 1}} = \frac{{\left( {x^4  - 1} \right) + 1}}<br />
{{x^2  + 1}} = \frac{{\left( {x^2  + 1} \right)\left( {x^2  - 1} \right) + 1}}<br />
{{x^2  + 1}} = x^2  - 1 + \frac{1}<br />
{{x^2  + 1}}.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Jun 2006
    From
    San Diego
    Posts
    101
    Quote Originally Posted by Krizalid View Post
    You don't even need partial fractions, you just need add & remove:

    \frac{{x^4 }}<br />
{{x^2  + 1}} = \frac{{\left( {x^4  - 1} \right) + 1}}<br />
{{x^2  + 1}} = \frac{{\left( {x^2  + 1} \right)\left( {x^2  - 1} \right) + 1}}<br />
{{x^2  + 1}} = x^2  - 1 + \frac{1}<br />
{{x^2  + 1}}.
    Much slicker!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    14
    I prefer this rather than partial fractions, it takes less time. (Of course, not always it's possible to avoid such technique.)
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Sep 2007
    Posts
    39
    Thanks alot guys! Kind of rediculous my teacher did not elaborate on this at all. Cheers! Kanpai! Konbae!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Quick integral question?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 28th 2011, 05:05 AM
  2. just a quick particular integral question..
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: May 22nd 2009, 08:11 AM
  3. quick integral question!
    Posted in the Calculus Forum
    Replies: 4
    Last Post: March 10th 2009, 01:14 PM
  4. Quick Conversion question
    Posted in the Math Topics Forum
    Replies: 17
    Last Post: September 1st 2008, 04:31 PM
  5. Quick integral question
    Posted in the Calculus Forum
    Replies: 3
    Last Post: June 13th 2008, 07:04 PM

Search Tags


/mathhelpforum @mathhelpforum