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Homework Help: Integrate in parts

  1. Jan 16, 2009 #1
    [tex]\int[/tex]x*arctg(1/x)dx

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

    by using
    [tex]\int[/tex]udv=uv-[tex]\int[/tex]vdu

    u=x
    du=dx

    dv=arctg(1/x)
    v=[1/(1+(1/x)2)]*ln|x|
    =[x2/1+x2]ln|x|

    [tex]\int[/tex]x*arctg(1/x)dx=x*arctg(1/x)-[tex]\int[/tex](x3ln|x|)/(x2+1)dx

    now.......
     
  2. jcsd
  3. Jan 16, 2009 #2

    Mark44

    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
    arctan(1/x)=u
    du=[1/(1 + (1/x)^2)]*ln|x|*dx

    xdx=dv
    v=0.5x^2

    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

    Mark44

    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]
     
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