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Re: Integral of x/(1+x^4) problem (in Mathematica30)

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  • Subject: [mg21702] Re: Integral of x/(1+x^4) problem (in Mathematica30)
  • From: "Kevin J. McCann" <kevin.mccann at>
  • Date: Sat, 22 Jan 2000 02:53:13 -0500 (EST)
  • Organization: Johns Hopkins University Applied Physics Lab, Laurel, MD, USA
  • References: <86aamp$>
  • Sender: owner-wri-mathgroup at

If you plot

 f[x_]=-(1/2) ArcTan[1/x^2]



you will see that they differ by Pi/4.  Also

Limit[f[x],{x->0}] = -Pi/4

The two answers differ by an integration constant, and are thus
equally good answers.  Depends on you criterion for "goodness".
I tend to agree with you on this, but mathematically it makes no



Kevin J. McCann
Johns Hopkins University APL

Jan Krupa <krupa at> wrote in message
news:86aamp$8ho at
> It is well known that antiderivative of x/(1+x^4) is
> 0.5*arctan(x^2) over the interval (-oo,oo).
> I tried it in mathematica3.0:
> Integrate[x/(1+x^4), x] gives
> -(1/2) ArcTan(1/x^2) which is correct but
> only for (-oo,0) or (0,oo).
> Of course one can define function F(x)= -(1/2) ArcTan(1/x^2) when x is
> not
> equal 0 and F(x)=0 when x=0.
> F(x)=0.5*ArcTan(x^2)+Pi/4 (because we have ArcTan(1/a)+ArcTan(a)=Pi/2
> for x not eq 0. )
> How one can force Mathematica to achieve  rather the better result
> 0.5*ArcTan(x^2) than -0.5*ArcTan(1/x^2) ?
> Jan
> P.S. MuPAD also gives -0.5*ArcTan(1/x^2).

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