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Math Help - The integral of 1/ 11+x^2

  1. #1
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    The integral of 1/ 11+x^2

    Problem:
    \int\frac{1}{11+x^2}dx

    Attempt:
    x = \sqrt{11}*tan(\theta)
    dx = \frac{\sqrt{11} *d\theta}{cos^2\theta}
    \int\frac{1}{11+(\sqrt{11}+tan(\theta))^2} * \frac{\sqrt{11} *d\theta}{cos^2\theta}

    \int\frac{\sqrt{11}*d\theta}{(11+\sqrt{11}*tan(\th  eta))^2*cos^2(\theta)}

    \int\frac{\sqrt{11}*d\theta}{(11+11*tan^2(\theta))  *cos^2(\theta)}

     <br />
\int\frac{\sqrt{11}*d\theta}{11(1+\frac{sin^2(\the  ta)}{cos^2(\theta})*cos^2(\theta)}<br />

     <br /> <br />
\int\frac{\sqrt{11}*d\theta}{11*(cos^2(\theta)+sin  ^2(\theta))}<br />

     <br /> <br />
\int\frac{\sqrt{11}*d\theta}{11}<br />
     <br /> <br />
\frac{\sqrt{11}*\theta}{11} + C<br />

    Answer:
    \frac{\sqrt{11}}{11}*arctan(\frac{x}{\sqrt{11}})+C

    Correct Answer:
    \frac{1}{\sqrt{11}}*arctan(\frac{x}{\sqrt{11}})+C

    Can someone help?
    Last edited by Fallen186; May 5th 2009 at 05:25 PM. Reason: I messed up on correct answer
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  2. #2
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    I don't see how the second line easily follows:

    <br /> <br />
\int\frac{\sqrt{11}*d\theta}{(11+11*tan^2(\theta))  *cos^2(\theta)}<br />

    Did you square the sum in the denominator? Maybe you just did a lot of simplification but I don't see how you got to the this line.

    edit: Well don't listen to me, skeeter says you're right, so I believe him. I didn't think you were necessarily wrong, I just didn't see the work between lines.
    Last edited by Jameson; May 5th 2009 at 04:39 PM.
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  3. #3
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    you are correct ... the "correct" answer is not.
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  4. #4
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    Quote Originally Posted by Jameson View Post
    I don't see how the second line easily follows:

    <br /> <br />
\int\frac{\sqrt{11}*d\theta}{(11+11*tan^2(\theta))  *cos^2(\theta)}<br />

    Did you square the sum in the denominator? Maybe you just did a lot of simplification but I don't see how you got to the this line.

    edit: Well don't listen to me, skeeter says you're right, so I believe him. I didn't think you were necessarily wrong, I just didn't see the work between lines.
    I simplified the dx into the big fraction and I foiled the (\sqrt{11}*tan(\theta))^2
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  5. #5
    Senior Member Pinkk's Avatar
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    Uhh... \frac{1}{\sqrt{11}}=\frac{\sqrt{11}}{11}. Those two answers are equal to one another.

    And I don't know if this has already been defined in your class, but:

    \int \frac{du}{a^{2}+u^{2}}=\frac{1}{a}tan^{-1}(\frac{u}{a})+C

    And for this problem, a=\sqrt{11},u=x.
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