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Math Help - Integration problem

  1. #1
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    Integration problem

    I'm working on a pretty long integration problem, and I've managed to split it up into simpler parts. One of them is:

    Integration problem-picture-27.png

    Simplifying further, I get:

    Integration problem-picture-28.png

    and

    Integration problem-picture-29.png

    But I'm stuck on the last part. Wolfram says arctan something, but I need to know how do it.
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  2. #2
    MHF Contributor harish21's Avatar
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    \displaystyle \int \dfrac{1}{b^2+x^2} \;dx = \dfrac {1}{b} tan^{-1}\bigg(\frac{x}{b}\bigg)+C
    Last edited by harish21; November 14th 2010 at 07:59 AM. Reason: typo
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  3. #3
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    Quote Originally Posted by TwoPlusTwo View Post
    I'm working on a pretty long integration problem, and I've managed to split it up into simpler parts. One of them is:

    Click image for larger version. 

Name:	Picture 27.png 
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ID:	19686

    Simplifying further, I get:

    Click image for larger version. 

Name:	Picture 28.png 
Views:	41 
Size:	6.8 KB 
ID:	19687

    and

    Click image for larger version. 

Name:	Picture 29.png 
Views:	40 
Size:	6.8 KB 
ID:	19688

    But I'm stuck on the last part. Wolfram says arctan something, but I need to know how do it.

    \displaystyle{\int\frac{1}{2+x^2}\,dx=\frac{1}{\sq  rt{2}}\int\frac{1/\sqrt{2}\,dx}{1+\left(\frac{x}{\sqrt{2}}\right)^2}  } \display{=\frac{1}{\sqrt{2}}\arctan\left(\frac{x}{  \sqrt{2}}\right)+C}

    Tonio
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  4. #4
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    Quote Originally Posted by harish21 View Post
    \displaystyle \int \dfrac{1}{b+x^2} \;dx = \dfrac {1}{b} tan^{-1}\bigg(\frac{x}{b}\bigg)+C
    That holds if and only if b = 1.

    \displaystyle \int \dfrac{1}{b+x^2} \;dx = \dfrac {1}{\sqrt{b}}\tan^{-1}\bigg(\frac{x}{\sqrt{b}}\bigg)+k.
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by TheCoffeeMachine View Post
    That holds if and only if b = 1.

    \displaystyle \int \dfrainc{1}{b+x^2} \;dx = \dfrac {1}{\sqrt{b}}\tan^{-1}\bigg(\frac{x}{\sqrt{b}}\bigg)+k.
    indeed. it's usually written \displaystyle \int \frac 1{b^2 + x^2}~dx = \frac 1b \tan ^{-1} \frac xb + C

    using \displaystyle a instead of \displaystyle b
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  6. #6
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    Correction already made by tonio [b]and[/b\] TheCoffeeMachine!
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