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Thread: Integral

  1. #1
    Senior Member polymerase's Avatar
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    Integral

    How do u integrate:

    $\displaystyle \displaystyle\int \dfrac{\sqrt{x-3}}{x^2}\,dx $

    Thanks
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by polymerase View Post
    How do u integrate:

    $\displaystyle \displaystyle\int \dfrac{\sqrt{x-3}}{x^2}\,dx $

    Thanks
    There might be an easier, more elegant, way to do this, but start with
    $\displaystyle x = 3 tan^2(\theta)$
    and see where that takes you.

    -Dan
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by polymerase View Post
    How do u integrate:

    $\displaystyle \displaystyle\int \dfrac{\sqrt{x-3}}{x^2}\,dx $
    Mmm. Integration by parts:

    $\displaystyle u=\sqrt{x-3}\implies du=\frac1{2\sqrt{x-3}}\,dx$ & $\displaystyle dv=x^{-2}\,dx\implies v=-\frac1x.$ The integral becomes to

    $\displaystyle \int {\frac{{\sqrt {x - 3} }}
    {{x^2 }}\,dx} = - \frac{{\sqrt {x - 3} }}
    {x} + \frac{1}
    {2}\underbrace {\int {\frac{1}
    {{x\sqrt {x - 3} }}\,dx} }_\varphi .$

    $\displaystyle \varphi$ is an arctangent, just set $\displaystyle \alpha=\sqrt{x-3}$ to descover it.
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  4. #4
    Senior Member polymerase's Avatar
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    Quote Originally Posted by Krizalid View Post
    Mmm. Integration by parts:

    $\displaystyle u=\sqrt{x-3}\implies du=\frac1{2\sqrt{x-3}}\,dx$ & $\displaystyle dv=x^{-2}\,dx\implies v=-\frac1x.$ The integral becomes to

    $\displaystyle \int {\frac{{\sqrt {x - 3} }}
    {{x^2 }}\,dx} = - \frac{{\sqrt {x - 3} }}
    {x} + \frac{1}
    {2}\underbrace {\int {\frac{1}
    {{x\sqrt {x - 3} }}\,dx} }_\varphi .$

    $\displaystyle \varphi$ is an arctangent, just set $\displaystyle \alpha=\sqrt{x-3}$ to descover it.
    huh? I get the byparts but the $\displaystyle \alpha$ stuff what?
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  5. #5
    Super Member PaulRS's Avatar
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    You have: $\displaystyle \int{\frac{dx}{x\cdot{2\cdot{\sqrt[]{x-3}}}}}$

    Now let: $\displaystyle u=\sqrt[]{x-3}$ then $\displaystyle \frac{du}{dx}=\frac{1}{2\cdot{\sqrt[]{x-3}}}$

    Thus we have: $\displaystyle \int{\frac{dx}{x\cdot{2\cdot{\sqrt[]{x-3}}}}}=\int{\frac{\left(\sqrt[]{x-3}\right)'}{x}\cdot{dx}}=\int{\frac{du}{u^2+3}}$

    Which is clearly of the arctan form

    $\displaystyle \int{\frac{du}{u^2+3}}=\frac{1}{3}\cdot{\int{\frac {du}{\left(\frac{u}{\sqrt[]{3}}\right)^2+1}}}=\frac{\sqrt[]{3}}{3}\cdot{\arctan (\frac{u}{\sqrt[]{3}}})+C$

    Which means that $\displaystyle \int{\frac{dx}{x\cdot{2\cdot{\sqrt[]{x-3}}}}}=\frac{\sqrt[]{3}}{3}\cdot{\arctan (\frac{\sqrt[]{x-3}}{\sqrt[]{3}}})+C$
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