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Math Help - \int \frac{x}{\sqrt{1+2x}}dx

  1. #1
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    \int \frac{x}{\sqrt{1+2x}}dx

    \int \frac{x}{\sqrt{1+2x}}dx
    Last edited by th%$&873; February 20th 2008 at 03:52 PM. Reason: title
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  2. #2
    Senior Member Peritus's Avatar
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    integration by parts:


    \int {\frac{x}<br />
{{\sqrt {1 + 2x} }}dx}  = x\sqrt {1 + 2x}  - \int {\sqrt {1 + 2x} dx = } x\sqrt {1 + 2x}  - \frac{1}<br />
{3}\left( {1 + 2x} \right)^{\frac{3}<br />
{2}}
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  3. #3
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    Quote Originally Posted by Peritus View Post
    integration by parts:


    \int {\frac{x}<br />
{{\sqrt {1 + 2x} }}dx}  = x\sqrt {1 + 2x}  - \int {\sqrt {1 + 2x} dx = } x\sqrt {1 + 2x}  - \frac{1}<br />
{3}\left( {1 + 2x} \right)^{\frac{3}<br />
{2}}
    Or make the sbstitution u = 1 + 2x \Rightarrow x = \frac{u-1}{2} \, and \, \frac{du}{dx} = 2 \Rightarrow dx = \frac{du}{2}.

    Then the integral becomes


    \int \frac{(u - 1)/2}{\sqrt{u}} \, \frac{du}{2} = \frac{1}{4} \int \frac{u - 1}{\sqrt{u}} \, du


    which can be broken up into simple bits involving powers of u.
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    mr fantastic, that's the method I used! And I got the right answer, but I made a small arithmetical error in the final simplification ( I added the denominators of 1/4 and 2/3 instead of multiplying them) so the answer didn't check. Sometimes I just see two numbers and instantly do the wrong operation.
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  5. #5
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    Quote Originally Posted by th%$&873 View Post
    mr fantastic, that's the method I used! And I got the right answer, but I made a small arithmetical error in the final simplification ( I added the denominators of 1/4 and 2/3 instead of multiplying them) so the answer didn't check. Sometimes I just see two numbers and instantly do the wrong operation.
    Became a pro at the abuse of substitution and you will save yourself a lot of headaches!
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  6. #6
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    Hello, th%$&873!

    I prefer to get rid of the radical first . . .


    \int \frac{x}{\sqrt{1+2x}}\,dx

    Let u \:= \:\sqrt{1+2x}\quad\Rightarrow\quad x \:=\: \frac{1}{2}(u^2-1)\quad\Rightarrow\quad dx \:=\:u\,du

    Substitute: . \int\frac{\frac{1}{2}(u^2-1)}{u}(u\,du) \;=\;\frac{1}{2}\int(u^2-1)\,du \;=\;\frac{1}{2}\left(\frac{u^3}{3} - u\right) + C \;=\;\frac{1}{6}\,u(u^2-3)+C

    Back-substitute: . \frac{1}{6}\sqrt{1+2x}\,(1 + 2x - 3) + C \;=\;\frac{1}{6}\sqrt{1+2x}\,(2x-2) + C

    . . Answer: . \frac{1}{3}\sqrt{1+2x}\,(x-1) + C

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  7. #7
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    Quote Originally Posted by th%$&873 View Post
    mr fantastic, that's the method I used! And I got the right answer, but I made a small arithmetical error in the final simplification ( I added the denominators of 1/4 and 2/3 instead of multiplying them) so the answer didn't check. Sometimes I just see two numbers and instantly do the wrong operation.
    Welcome to the club, chum!
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