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Math Help - Partial Fraction Integral

  1. #1
    Junior Member
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    Partial Fraction Integral

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

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

    <br />
-4\int\frac{1}{(x^2+4)^2}~dx<br />


    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:

    <br />
-4\bigg(\frac{x}{8(x^2+4)}+\frac{1}{16}arctan\bigg[\frac{x}{2}\bigg]\bigg)<br />

    I'm in a 5 week Calc2 course and I feel like this guy --> after just the first week.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Pennsylvania
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    Quote Originally Posted by tcRom View Post
    Here's the problem, I keep getting stuck on one part...
    <br />
\int\frac{x^2+2x}{(x^2+4)^2}~dx<br />

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

    <br />
-4\int\frac{1}{(x^2+4)^2}~dx<br />


    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:

    <br />
-4\bigg(\frac{x}{8(x^2+4)}+\frac{1}{16}arctan\bigg[\frac{x}{2}\bigg]\bigg)<br />

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

    Let x=2\tan(\theta)

    Try that
    Last edited by Mathstud28; July 19th 2008 at 06:27 PM.
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  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...
    <br />
\int\frac{x^2+2x}{(x^2+4)^2}~dx<br />

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

    <br />
-4\int\frac{1}{(x^2+4)^2}~dx<br />


    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:

    <br />
-4\bigg(\frac{x}{8(x^2+4)}+\frac{1}{16}arctan\bigg[\frac{x}{2}\bigg]\bigg)<br />

    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

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

    And here is what we do

    Let

    x=2\tan(\theta)

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

    So we have

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

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

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

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

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

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

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

    \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

    =\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

    =\frac{1}{16}\bigg[\arctan\left(\frac{x}{2}\right)+\frac{2x}{x^2+4}\b  igg]+C
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  4. #4
    Junior Member
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    It's apparent that I've stepped into your kingdom.

    Thank you very much for the help!
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