Results 1 to 7 of 7

Thread: \int \frac{x}{\sqrt{1+2x}}dx

  1. #1
    Newbie
    Joined
    Dec 2007
    Posts
    19

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

    $\displaystyle \int \frac{x}{\sqrt{1+2x}}dx$
    Last edited by th%$&873; Feb 20th 2008 at 03:52 PM. Reason: title
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member Peritus's Avatar
    Joined
    Nov 2007
    Posts
    397
    integration by parts:


    $\displaystyle \int {\frac{x}
    {{\sqrt {1 + 2x} }}dx} = x\sqrt {1 + 2x} - \int {\sqrt {1 + 2x} dx = } x\sqrt {1 + 2x} - \frac{1}
    {3}\left( {1 + 2x} \right)^{\frac{3}
    {2}} $
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    Quote Originally Posted by Peritus View Post
    integration by parts:


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

    Then the integral becomes


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


    which can be broken up into simple bits involving powers of u.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Dec 2007
    Posts
    19
    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.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    GAMMA Mathematics
    colby2152's Avatar
    Joined
    Nov 2007
    From
    Alexandria, VA
    Posts
    1,172
    Thanks
    1
    Awards
    1
    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!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    Hello, th%$&873!

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


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

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

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

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

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

    Follow Math Help Forum on Facebook and Google+

  7. #7
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    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!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: Mar 26th 2011, 06:05 AM
  2. Replies: 3
    Last Post: Mar 26th 2011, 05:43 AM
  3. \int{\frac{x^2}{\sqrt{4x-x^2}}}dx
    Posted in the Calculus Forum
    Replies: 5
    Last Post: Feb 21st 2008, 06:15 AM
  4. Replies: 11
    Last Post: Jan 6th 2008, 09:33 AM
  5. prove sqrt(3) + sqrt (5) is an irrational number
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: Oct 6th 2006, 06:48 PM

Search Tags


/mathhelpforum @mathhelpforum