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Math Help - Integration

  1. #1
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    Integration

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  2. #2
    Senior Member Peritus's Avatar
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    <br />
\int {\frac{{4x}}<br />
{{\sqrt {2x - 1} }}} dx = 2\int {\frac{{2x + 1 - 1}}<br />
{{\sqrt {2x - 1} }}} dx = 2\int {\sqrt {2x - 1}  + 2\int {\frac{1}<br />
{{\sqrt {2x - 1} }}} } dx<br />

    and now the result follows immediately:

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

    <br />
 = \sqrt {2x - 1} \left( {\frac{4}<br />
{3}x - \frac{2}<br />
{3} + 2} \right) + C = \frac{4}<br />
{3}\sqrt {2x - 1} (x - 1) + C<br />
    Last edited by Peritus; January 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 u^2=2x-1\implies u\,du=dx, the integral becomes

    2\int {\left( {u^2  + 1} \right)\,du}  = 2\left( {\frac{1}<br />
{3}u^3  + u} \right) + k = 2u\left( {\frac{1}<br />
{3}u^2  + 1} \right) + k.

    Back substitute:

    \int {\frac{{4x}}<br />
{{\sqrt {2x - 1} }}\,dx}  = 2\sqrt {2x - 1} \left( {\frac{{2x - 1}}<br />
{3} + 1} \right) = \frac{4}<br />
{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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