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Integrating ln(x+1)/(x^2+1) using recursive integration by parts

  1. May 6, 2012 #1
    Hi, I need to find ∫ln(x+1)/(x^2+1)dx

    I think it might involve recursive integration by parts, so first I set:

    u=ln(x+1) dv = 1/(x^2+1)dx
    du=1/(x+1)dx v=ArcTan(x)

    ∫ln(x+1)/(x^2+1)dx = ArcTan(x)Ln(x+1) - ∫ArcTan(x)/(x+1)dx

    Then I integrated by parts again, so

    u=1/(x+1) dv=ArcTan(x)
    du=-1/(x+1)^2dx v=x*ArcTan(x)-Ln(x^2+1)/(x+1)

    ∫Ln(x+1)/(x^2+1)dx = ArcTan(x)Ln(x+1) -1/(x+1)(x*ArcTan(x)-Ln(x^2+1)/2)+∫x*ArcTan(x)/(x+1)^2-Ln(x^2+1)/(2(x+1)^2)

    my problem here is that neither my ArcTan term nor my natural log term that I'm attempting to integrate can be rewritten in terms of either of the two previous integrals I've done. I have the natural log of (x^2+1) instead of (x+1) and my ArcTan term is divided by (x+1)^2 rather than (x+1). So I guess I could continue integration by parts, but that's just getting real ugly real fast. Any ideas?
     
  2. jcsd
  3. May 7, 2012 #2
    Hi,
    This indefinite integral cannot be expressed with a combination of a finite number of the usual functions. A formal expression (a closed form) of the integral involves special functions of the polylogarithm kind.
    Nevertheless, il the limits of integration were specified, in some particular cases it might be possible to express the integral as a combination of a finite number of the usual functions.
     
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