Results 1 to 5 of 5

Math Help - Improper Integration

  1. #1
    Member >_<SHY_GUY>_<'s Avatar
    Joined
    Jan 2008
    From
    CA
    Posts
    181

    Question Improper Integration

    The integral from 1 to infinity [sorry i don't know how to put it in LaTex ], of:
    \frac {3dx}{3x-x^2}

    From what My teacher showed me, you can use Ln, or partial fractions to solve this, but I couldn't understand either method, how do you do it?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by >_<SHY_GUY>_< View Post
    The integral from 1 to infinity [sorry i don't know how to put it in LaTex ], of:
    \frac {3dx}{3x-x^2}

    From what My teacher showed me, you can use Ln, or partial fractions to solve this, but I couldn't understand either method, how do you do it?
    Resolve into partial fractions:

    \frac{3}{3x-x^2}=\frac{A}{x}+\frac{B}{3-x}

    and a quick calculation shows that:

    \frac{3}{3x-x^2}=\frac{1}{x}+\frac{1}{3-x}

    Which has resolved the integrand into the sum of two terms both of which are easily integrable (in terms of \ln )

    CB
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member >_<SHY_GUY>_<'s Avatar
    Joined
    Jan 2008
    From
    CA
    Posts
    181
    Quote Originally Posted by CaptainBlack View Post
    Resolve into partial fractions:

    \frac{3}{3x-x^2}=\frac{A}{x}+\frac{B}{3-x}

    and a quick calculation shows that:

    \frac{3}{3x-x^2}=\frac{1}{x}+\frac{1}{3-x}

    Which has resolved the integrand into the sum of two terms both of which are easily integrable (in terms of \ln )

    CB
    from there you solve for A and B and then take the integrals of those fractions...Im on the right track right?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1
    Yes. In the second step CB solved for A, B and found A=B=1. Now you can integrate both fractions separately, which is easy.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Nov 2009
    Posts
    263
    Quote Originally Posted by >_<SHY_GUY>_< View Post
    from there you solve for A and B and then take the integrals of those fractions...Im on the right track right?
    yes,..go on
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Improper Integration
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 9th 2011, 11:31 PM
  2. Improper Integration
    Posted in the Calculus Forum
    Replies: 6
    Last Post: April 17th 2010, 09:17 PM
  3. improper integration by parts
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: December 9th 2009, 05:52 AM
  4. Improper Integration
    Posted in the Calculus Forum
    Replies: 9
    Last Post: June 22nd 2008, 12:33 PM
  5. improper integration
    Posted in the Calculus Forum
    Replies: 3
    Last Post: July 15th 2007, 06:52 PM

Search Tags


/mathhelpforum @mathhelpforum