Results 1 to 4 of 4

Thread: Partial Fraction Integral

  1. #1
    Junior Member
    Joined
    Jul 2008
    Posts
    38

    Partial Fraction Integral

    Here's the problem, I keep getting stuck on one part...
    $\displaystyle
    \int\frac{x^2+2x}{(x^2+4)^2}~dx
    $

    After separating into 2 integrals and taking the partial fraction of the first part, I end up getting stuck on the following integral:

    $\displaystyle
    -4\int\frac{1}{(x^2+4)^2}~dx
    $


    My question at this point: what's my substitution? I'm not seeing it but know, based on form, that it has something to do with arctan. I picked up the answer from the Mathematica Integrator but need to know the steps.

    The answer is apparently:

    $\displaystyle
    -4\bigg(\frac{x}{8(x^2+4)}+\frac{1}{16}arctan\bigg[\frac{x}{2}\bigg]\bigg)
    $

    I'm in a 5 week Calc2 course and I feel like this guy --> after just the first week.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by tcRom View Post
    Here's the problem, I keep getting stuck on one part...
    $\displaystyle
    \int\frac{x^2+2x}{(x^2+4)^2}~dx
    $

    After separating into 2 integrals and taking the partial fraction of the first part, I end up getting stuck on the following integral:

    $\displaystyle
    -4\int\frac{1}{(x^2+4)^2}~dx
    $


    My question at this point: what's my substitution? I'm not seeing it but know, based on form, that it has something to do with arctan. I picked up the answer from the Mathematica Integrator but need to know the steps.

    The answer is apparently:

    $\displaystyle
    -4\bigg(\frac{x}{8(x^2+4)}+\frac{1}{16}arctan\bigg[\frac{x}{2}\bigg]\bigg)
    $

    I'm in a 5 week Calc2 course and I feel like this guy --> after just the first week.

    Let $\displaystyle x=2\tan(\theta)$

    Try that
    Last edited by Mathstud28; Jul 19th 2008 at 06:27 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by tcRom View Post
    Here's the problem, I keep getting stuck on one part...
    $\displaystyle
    \int\frac{x^2+2x}{(x^2+4)^2}~dx
    $

    After separating into 2 integrals and taking the partial fraction of the first part, I end up getting stuck on the following integral:

    $\displaystyle
    -4\int\frac{1}{(x^2+4)^2}~dx
    $


    My question at this point: what's my substitution? I'm not seeing it but know, based on form, that it has something to do with arctan. I picked up the answer from the Mathematica Integrator but need to know the steps.

    The answer is apparently:

    $\displaystyle
    -4\bigg(\frac{x}{8(x^2+4)}+\frac{1}{16}arctan\bigg[\frac{x}{2}\bigg]\bigg)
    $

    I'm in a 5 week Calc2 course and I feel like this guy --> after just the first week.
    Sorry, I had to shave .

    Ok, so here is what we have

    $\displaystyle \int\frac{dx}{(x^2+4)^2}$

    And here is what we do

    Let

    $\displaystyle x=2\tan(\theta)$

    $\displaystyle \Rightarrow{dx=2\sec^2(\theta)~d\theta}$

    So we have

    $\displaystyle \int\frac{dx}{(x^2+4)^2}$

    $\displaystyle \underbrace{\Rightarrow}_{x=2\tan(\theta)}\int\fra c{2\sec^2(\theta)}{(4\tan^2(\theta)+4)^2}~d\theta$

    $\displaystyle =\frac{1}{8}\int\frac{\sec^2(\theta)}{\sec^4(\thet a)}~d\theta$

    $\displaystyle =\frac{1}{8}\int\cos^2(\theta)~d\theta$

    $\displaystyle =\frac{1}{16}\int\bigg[1+\cos(2\theta)\bigg]~d\theta$

    $\displaystyle =\frac{1}{16}\bigg[\theta+\frac{1}{2}\sin(2\theta)\bigg]+C$

    $\displaystyle =\frac{1}{16}\bigg[\theta+\sin(\theta)\cos(\theta)\bigg]$

    $\displaystyle \underbrace{=}_{\text{backsub}}\frac{1}{16}\bigg[\arctan\left(\frac{x}{2}\right)+\sin\left(\arctan\ left(\frac{x}{2}\right)\right)\cos\left(\arctan\le ft(\frac{x}{2}\right)\right)\bigg]+C$

    $\displaystyle =\frac{1}{16}\bigg[\arctan\left(\frac{x}{2}\right)+\frac{x}{\sqrt{x^2 +4}}\cdot\frac{2}{\sqrt{x^2+4}}\bigg]+C$

    $\displaystyle =\frac{1}{16}\bigg[\arctan\left(\frac{x}{2}\right)+\frac{2x}{x^2+4}\b igg]+C$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jul 2008
    Posts
    38
    It's apparent that I've stepped into your kingdom.

    Thank you very much for the help!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Integral with Partial Fraction Decomposition
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Sep 13th 2011, 03:49 PM
  2. Big Integral (partial fraction decomposition?)
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Nov 16th 2010, 10:16 PM
  3. help with partial fraction integral
    Posted in the Calculus Forum
    Replies: 6
    Last Post: Apr 28th 2009, 11:54 AM
  4. Integral - Partial Fraction
    Posted in the Calculus Forum
    Replies: 7
    Last Post: Feb 19th 2008, 04:23 PM
  5. integral (partial fraction)
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Jun 13th 2007, 08:56 PM

Search Tags


/mathhelpforum @mathhelpforum