1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
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 product

  1. Mar 19, 2009 #1
    In the following problem where does the du/2 come from?
    I don't understand why it isn't just du.
    http://users.on.net/~rohanlal/partialdif.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 19, 2009 #2

    danago

    User Avatar
    Gold Member

    They have made the substitution u = x^2+1, hence the corrosponding differential du is given by:

    du = 2x dx

    Hence, after rearranging:

    dx = du / 2x

    Sub this back into the integral, and the x's will cancel out, leaving du/2.
     
  4. Mar 19, 2009 #3
    As a calculus student myself, the method we use is slightly different (and I think more commonly taught).

    Instead of keeping the 1/2 in the integral, since it is a constant, pull it out to the front of the entire integral, complete the integral, then multiply it by the 1/2 again.

    So [tex]\int x(x^2+1)^{-1} dx[/tex]
    [tex]u=x^2+1, du=2xdx[/tex] Now at this step, divide both sides by 2 (which is what they originally did in the problem, but instead of just leaving du/2, leave it has [tex]\frac{1}{2} \int \frac{du}{u}[/tex]

    This is a much simpler notation, since you now have a simple du/u to integrate.

    Integrate, [tex]ln(x^2+1)[/tex] then multiply by 1/2!

    [tex]\frac{1}{2} ln(x^2+1)[/tex] This also makes the logarithmic operation of raising to the 1/2 easier to spot.

    Again, all this is quite trivial as it has no actual effect on the practice of integration. But it IS an easier notation to work with (or so I believe).
     
    Last edited: Mar 19, 2009
  5. Mar 24, 2009 #4
    when differentiating u why does it gain a dx?
    dx is usually only present when attempting to find the integral of something, not when stating the derivative of a function.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook