Re: Question about an integral

*To*: mathgroup at smc.vnet.net*Subject*: [mg54455] Re: [mg54381] Question about an integral*From*: DrBob <drbob at bigfoot.com>*Date*: Sun, 20 Feb 2005 00:11:13 -0500 (EST)*References*: <200502190732.CAA06170@smc.vnet.net>*Reply-to*: drbob at bigfoot.com*Sender*: owner-wri-mathgroup at wolfram.com

Integrate[1/x^2, {x, 2, A}] (-2 + A)*If[Re[A] >= 0 || Im[A] != 0, 1/(2*A), Integrate[1/(2 + (-2 + A)*x)^ 2, {x, 0, 1}, Assumptions -> !(Re[A] >= 0 || Im[A] != 0)]] Simplify[%, A > 0] 1/2 - 1/A That's the wrong answer if A < 0, at least, so it's not odd if Mathematica doesn't return it for unknown A. Bobby On Sat, 19 Feb 2005 02:32:53 -0500 (EST), richard speers <rspeers at skidmore.edu> wrote: > The previous version of Mathematica integrated 1/x^2 from x=2 to x=A > very nicely, giving 1/2-1/A. The new > version won't do the integral. Please help. > > > > -- DrBob at bigfoot.com www.eclecticdreams.net

**References**:**Question about an integral***From:*richard speers <rspeers@skidmore.edu>