1. Not finding help here? Sign up for a free 30min 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!

Integrate in parts

  1. Jan 16, 2009 #1

    i have such problems integrating arc functions, dont know why, but they never turn out right,,

    by using




  2. jcsd
  3. Jan 16, 2009 #2


    Staff: Mentor

    Your work going from dv to v is incorrect.
    You have dv = arctan(1/x) (minor quibble: it should be arctan(1/x)dx), but from that you can't get to v = [1/(1 + (1/x)^2)]*ln|x|.

    A possibility when you're doing integration by parts is arriving at an integral that's harder than the one you started with. That's usually a sign that what you picked for u and dv is not the right choice.

    For this problem, I believe that the right choice is

    u = arctan(1/x), dv = xdx
  4. Jan 17, 2009 #3
    du=[1/(1 + (1/x)^2)]*ln|x|*dx


    is this correct?
    this will also give me an integration harder than i started with

    [tex]\int[/tex]x*arctg(1/x)dx=0.5x^2arctan(1/x)-[tex]\int[/tex][0.5x^2/(1 + (1/x)^2)]*ln|x|dx
  5. Jan 17, 2009 #4


    Staff: Mentor

    For du, you're close, but you have the antiderivative of 1/x rather than its derivative. IOW, you have d/dx(1/x) = ln|x|, which is wrong.
    Starting with u = arctan(1/x), and letting w = 1/x, we have
    u = arctan(w).
    So du = d/dw(arctan(w)) *dw = d/dw(arctan(w)) * dw/dx * dx
    = 1/[1 + w^2] * (-1)/x^2 * dx (the factor before dx is where you made your mistake.)
    = 1/[1 + (1/x)^2]* (-1)/x^2 * dx
    [tex]\frac{-dx}{(1 + (1/x)^2)x^2}[/tex]
    = [tex]\frac{-dx}{x^2 + 1}[/tex]
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Integrate in parts
  1. Integration by parts (Replies: 6)

  2. Integration by parts (Replies: 8)

  3. Integration by Parts (Replies: 8)