Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Integral of 1/1+sqrt(x)

  1. Oct 12, 2009 #1
    I'm having trouble calculating an integral:

    [tex]\int_0^1{\frac{1}{1+\sqrt{x}}dx}[/tex]

    I decided to do a substitution:

    [tex]u=\sqrt{x}[/tex]
    [tex]du=\frac{1}{2\sqrt{x}}dx=\frac{1}{2u}dx[/tex]
    thus making the integral look like this:
    [tex]\int_0^1{\frac{1}{2u}\cdot\frac{1}{1+u}du}[/tex]
    I transformed this integral to:
    [tex]\int_0^1{\frac{1}{2u}\cdot\frac{1}{1+u}du}=\int_0^1{\frac{A}{2u}+\frac{B}{1+u}du=\int_0^1{\frac{1}{2u}+\frac{(-1/2)}{1+u}du[/tex]
    And tried to solve the integral like this:
    [tex]\ln|2u|-\frac{1}{2}\ln|1+u|[/tex]
    going from 0 to 1.

    However, this means to insert 0 in an ln-expression. Obviously I have done something wrong? Can someone please help?
     
  2. jcsd
  3. Oct 12, 2009 #2
    Nevermind. I did it now.

    It was a simple mistake on my part.

    The integral should look like:

    [tex]\int_0^1{\frac{2u}{1+u}du}[/tex]

    after substitution

    and then it should be calculated with integration by parts from there.
     
  4. Oct 12, 2009 #3
    [tex]\left.{\frac{1}{1+\sqrt{x}}\right|_{x=1} = \left\{\frac{1}{2},\infty\right\}[/tex]
     
  5. Oct 12, 2009 #4

    Gib Z

    User Avatar
    Homework Helper

    I'm not entirely sure what that post means trambolin, care to explain?

    Dragonblood - The solution comes out must quicker than integration by parts if you add and subtract 2 into the numerator.
     
  6. Oct 12, 2009 #5
    I meant the -1 is causing the integrand blow up at x=1 for the cause of ln 0. But nevermind it is just the notation stuff...
     
  7. Oct 12, 2009 #6
    Ah, yes. That is much easier :) Thanks
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Integral of 1/1+sqrt(x)
  1. Integral: x/1+sqrt(x) (Replies: 7)

Loading...