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

• To: mathgroup at smc.vnet.net
• Subject: [mg21674] Integral of x/(1+x^4) problem (in Mathematica30)
• From: Jan Krupa <krupa at alpha.sggw.waw.pl>
• Date: Fri, 21 Jan 2000 04:00:33 -0500 (EST)
• Organization: http://news.icm.edu.pl/
• Sender: owner-wri-mathgroup at wolfram.com

```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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