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Thread: Integration

  1. #1
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    Integration

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  2. #2
    Senior Member Peritus's Avatar
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    $\displaystyle
    \int {\frac{{4x}}
    {{\sqrt {2x - 1} }}} dx = 2\int {\frac{{2x + 1 - 1}}
    {{\sqrt {2x - 1} }}} dx = 2\int {\sqrt {2x - 1} + 2\int {\frac{1}
    {{\sqrt {2x - 1} }}} } dx
    $

    and now the result follows immediately:

    $\displaystyle
    = {\raise0.7ex\hbox{$2$} \!\mathord{\left/
    {\vphantom {2 3}}\right.\kern-\nulldelimiterspace}
    \!\lower0.7ex\hbox{$3$}}\left( {2x - 1} \right)^{{\raise0.7ex\hbox{$3$} \!\mathord{\left/
    {\vphantom {3 2}}\right.\kern-\nulldelimiterspace}
    \!\lower0.7ex\hbox{$2$}}} + 2\sqrt {2x - 1} + C

    $

    $\displaystyle
    = \sqrt {2x - 1} \left( {\frac{4}
    {3}x - \frac{2}
    {3} + 2} \right) + C = \frac{4}
    {3}\sqrt {2x - 1} (x - 1) + C
    $
    Last edited by Peritus; Jan 21st 2008 at 01:34 PM.
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    I'd tackle this as follows:

    Substitute $\displaystyle u^2=2x-1\implies u\,du=dx,$ the integral becomes

    $\displaystyle 2\int {\left( {u^2 + 1} \right)\,du} = 2\left( {\frac{1}
    {3}u^3 + u} \right) + k = 2u\left( {\frac{1}
    {3}u^2 + 1} \right) + k.$

    Back substitute:

    $\displaystyle \int {\frac{{4x}}
    {{\sqrt {2x - 1} }}\,dx} = 2\sqrt {2x - 1} \left( {\frac{{2x - 1}}
    {3} + 1} \right) = \frac{4}
    {3}(x + 1)\sqrt {2x - 1} + k\quad\blacksquare$

    ---

    Peritus, I'd re-check your last conclusion.
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  4. #4
    Senior Member Peritus's Avatar
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    Peritus, I'd re-check your last conclusion
    it's not even a conclusion it'ss pure bull@#@. thanks for pointing out my mistake. I've corrected my post.
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