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Math Help - Integral

  1. #1
    Senior Member polymerase's Avatar
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    Integral

    How do u integrate:

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

    Thanks
    There might be an easier, more elegant, way to do this, but start with
    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\int \dfrac{\sqrt{x-3}}{x^2}\,dx
    Mmm. Integration by parts:

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

    \int {\frac{{\sqrt {x - 3} }}<br />
{{x^2 }}\,dx} = - \frac{{\sqrt {x - 3} }}<br />
{x} + \frac{1}<br />
{2}\underbrace {\int {\frac{1}<br />
{{x\sqrt {x - 3} }}\,dx} }_\varphi .

    \varphi is an arctangent, just set \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:

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

    \int {\frac{{\sqrt {x - 3} }}<br />
{{x^2 }}\,dx} = - \frac{{\sqrt {x - 3} }}<br />
{x} + \frac{1}<br />
{2}\underbrace {\int {\frac{1}<br />
{{x\sqrt {x - 3} }}\,dx} }_\varphi .

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

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

    Thus we have: \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

    \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 \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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