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Re: Integral of x/(1+x^4) problem (in Mathematica30)
*To*: mathgroup at smc.vnet.net
*Subject*: [mg21695] Re: [mg21674] Integral of x/(1+x^4) problem (in Mathematica30)
*From*: Andrzej Kozlowski <andrzej at tuins.ac.jp>
*Date*: Sat, 22 Jan 2000 02:52:58 -0500 (EST)
*Sender*: owner-wri-mathgroup at wolfram.com
You can get close to what you want simply by:
In[1]:=
FullSimplify[Integrate[x/(1+x^4), x]]
Out[1]=
1 2
-(-) ArcCot[x ]
2
This is differs from the answer you wanted by Pi/2 and is also defined over
the whole real axis.
> From: Jan Krupa <krupa at alpha.sggw.waw.pl>
> Organization: http://news.icm.edu.pl/
> Date: Fri, 21 Jan 2000 04:00:33 -0500 (EST)
> To: mathgroup at smc.vnet.net
> Subject: [mg21695] [mg21674] Integral of x/(1+x^4) problem (in Mathematica30)
>
> 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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