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) ? >