Results 1 to 4 of 4

Math Help - integral question

  1. #1
    Newbie
    Joined
    Jul 2010
    Posts
    4

    integral question

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

    according to my book it says let u=1+\sqrt{2x}

    du=\frac{1}{\sqrt{2x}}dx-->(u-1)du=dx

    I do not understand how du=\frac{1}{\sqrt{2x}}dx-->(u-1)du=dx

    when i try to take the derivative of U i get: du= \frac{d}{dx}[1+\sqrt{2x}]=\frac{2}{3}(2x)^\frac{3}{2}dx

    Im having problems getting from here: \int\frac{1}{1+\sqrt{2x}} to here \int\frac{(u-1)}{u}du


    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Unknown008's Avatar
    Joined
    May 2010
    From
    Mauritius
    Posts
    1,260
        du=\frac{1}{\sqrt{2x}}dx

    This becomes:

     \sqrt{2x}   du=dx

    What is \sqrt{2x}?

    u = 1 + \sqrt{2x}

    u - 1 = \sqrt{2x}

    Hence;

     \sqrt{2x}   du=dx

     (u-1)du=dx
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Unknown008's Avatar
    Joined
    May 2010
    From
    Mauritius
    Posts
    1,260
    du= \dfrac{d}{dx}[1+\sqrt{2x}]=\frac{1}{2}(2x)^{-\frac{1}{2}}dx

    This is what you should do. You integrated it instead of differentiating it.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jul 2010
    Posts
    4
    Thats it.. chain rule... got it.. thanks a lot..
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Integral question
    Posted in the Calculus Forum
    Replies: 4
    Last Post: March 28th 2011, 09:00 AM
  2. Double Integral and Triple Integral question
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 3rd 2010, 12:47 PM
  3. Integral Question
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 16th 2009, 11:38 AM
  4. Integral question:
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 25th 2009, 04:25 PM
  5. integral question
    Posted in the Calculus Forum
    Replies: 6
    Last Post: November 14th 2006, 01:26 PM

Search Tags


/mathhelpforum @mathhelpforum