Results 1 to 8 of 8

Math Help - Integration

  1. #1
    Newbie
    Joined
    Apr 2011
    Posts
    10

    Integration

    Hi there can anyone help me integrate the following
    \int (x^2+1)/(x^4+1)

    I've been trying unsuccessfully
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    What have you tried?

    Maybe break it up using partial fractions?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Mar 2010
    Posts
    715
    Thanks
    2
    I = \int\frac{x^2+1}{x^4+1}\;{dx} = \int\frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}}\;{dx  } = \int\frac{1+\frac{1}{x^2}}{(x-\frac{1}{x})^2+2}\;{dx}. Let t = x-1/x, then:

    I = \int\frac{1}{t^2+2}\;{dt} which hopefully is an integral that you are familiar with! :)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Apr 2011
    Posts
    10
    Ya that's what I've been trying I've x^4+1 factored as follows
    (x^2+(2^1/2)x+1)(x^2-(2^1/2)x+1)
    I'm having trouble solving the partial fractions though. Any ideas?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,513
    Thanks
    1404
    x^4 + 1 doesn't factorise... You should follow TheCoffeeMachine's method.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Apr 2011
    Posts
    10
    Great thanks ya I know that one inverse tan over root 2 all over root 2 right??!!! Thanks again guys really helpful site!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Mar 2010
    Posts
    715
    Thanks
    2
    Quote Originally Posted by userod View Post
    Ya that's what I've been trying I've factored as follows
    (x^2+(2^1/2)x+1)(x^2-(2^1/2)x+1)
    I'm having trouble solving the partial fractions though. Any ideas?
    Yes (I'm full of ideas this morning, haha), note that (x^2+√2x+1)+(x^2-√2x+1) = 2(x^2+1), so that:

    \begin{aligned} I & = \int\frac{x^2+1}{(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)}\;{dx} = \frac{1}{2}\int\frac{(x^2+\sqrt{2}x+1)+(x^2-\sqrt{2}x+1)}{(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)}\;{dx} \\& = \frac{1}{2}\int\frac{1}{x^2-\sqrt{2}x+1}\;{dx}+\frac{1}{2}\int\frac{1}{x^2+ \sqrt{2}x+1}\;{dx} = I_{1}+I_{2}, ~ \text{say}. \end{aligned}


    I_{1} = \frac{1}{2}\int\frac{1}{\left(x-\frac{1}{\sqrt{2}}\right)^2+\left(\frac{1}{\sqrt{2  }}\right)^2}\;{dx} = \frac{1}{\sqrt{2}}\tan^{-1}\left(\sqrt{2}x-1\right)+k_{1} and

    I_{2} = \frac{1}{2}\int\frac{1}{\left(x+\frac{1}{\sqrt{2}}  \right)^2+\left(\frac{1}{\sqrt{2}}\right)^2}\;{dx} = \frac{1}{\sqrt{2}}\tan^{-1}\left(\sqrt{2}x+1\right)+k_{2}, thus:

    I = \frac{1}{\sqrt{2}}\tan^{-1}\left(\sqrt{2}x-1\right)+ \frac{1}{\sqrt{2}}\tan^{-1}\left(\sqrt{2}x+1\right)+k.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by pickslides View Post
    What have you tried?

    Maybe break it up using partial fractions?
    Noting of course that x^4 + 1 = (x^4 + 2x^2 + 1) - 2x^2 = (x^2 + 1)^2 - 2x^2 = (x^2 + 1 - \sqrt{2}x)(x^2 + 1 + \sqrt{2}x).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: November 3rd 2010, 12:54 AM
  2. Replies: 2
    Last Post: November 2nd 2010, 04:57 AM
  3. Replies: 8
    Last Post: September 2nd 2010, 12:27 PM
  4. Replies: 2
    Last Post: February 19th 2010, 10:55 AM
  5. Replies: 6
    Last Post: May 25th 2009, 06:58 AM

Search Tags


/mathhelpforum @mathhelpforum