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Integration by Parts: Int: x*arctan(x) dx

  1. May 27, 2009 #1
    Because of circumstance (my desire to graduate in 5 years or less), I've been forced to attempt Calc 2 in 2 months time online over the summer. About 75% of it is going smoothly (compared with 105% or so of Calc 1).

    1. The problem statement, all variables and given/known data

    I'm to solve the indefinite integral: [tex]\int[/tex] x * arctan(x) dx



    2. Relevant equations
    Integration by parts is done using: [tex]\int[/tex]u dv = uv - [tex]\int[/tex]v du


    3. The attempt at a solution
    It seems pretty obvious that u = arctan(x) and that dv = x. From this, du = [tex]\frac{1}{1+x^2}[/tex] dx and v = [tex]\frac{1}{2}[/tex]x2.

    Using the integration by parts formula:

    [tex]\int[/tex] x * arctan(x) dx = [tex]\frac{1}{2}[/tex]x2arctan(x) - [tex]\frac{1}{2}[/tex][tex]\int[/tex][tex]\frac{x^2}{1 + x^2}[/tex]dx



    Now integration by parts must be used again. It seems obvious to select u = x2, dv = [tex]\frac{1}{1 + x^2}[/tex]. du = 2xdx, dv = arctan(x).



    [tex]\int[/tex] x * arctan(x) dx = [tex]\frac{1}{2}[/tex]x2arctan(x) - [tex]\frac{1}{2}[/tex] ( x2arctan(x) - 2 [tex]\int[/tex]xarctan(x) ).

    Simplified:

    [tex]\int[/tex] x * arctan(x) dx = 0 + [tex]\int[/tex] x * arctan(x) dx.



    While this is very true, it doesn't help me find the integral. Switching my u and dv in either use of the integration by parts formula hasn't yielded a solution for me in my attempts yet. Thank you in advance for your help :smile:.
     
  2. jcsd
  3. May 27, 2009 #2

    rock.freak667

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    Homework Helper

    [tex]\frac{x^2}{x^2+1}=\frac{x^2+1 -1}{x^2+1}=1-\frac{1}{x^2+1}[/tex]
     
  4. May 27, 2009 #3
    You are my Bokonon, only you tell truths.
     
  5. May 27, 2009 #4
    Thank you, in other words.
     
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