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

• To: mathgroup at smc.vnet.net
• Subject: [mg21690] Re: [mg21674] Integral of x/(1+x^4) problem (in Mathematica30)
• From: BobHanlon at aol.com
• Date: Sat, 22 Jan 2000 02:52:49 -0500 (EST)
• Sender: owner-wri-mathgroup at wolfram.com

```Use FullSimplify

Integrate[x/(1 + x^4), x]

-(1/2)*ArcTan[1/x^2]

Plot[%, {x, -5, 5}];

FullSimplify[%%]

-(1/2)*ArcCot[x^2]

Plot[%, {x, -5, 5}];

Note

Plot[1/2 * ArcTan[x^2], {x, -5, 5}];

Bob Hanlon

In a message dated 1/21/2000 5:32:00 AM, krupa at alpha.sggw.waw.pl writes:

>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) ?
>

```

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