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Re: Question about an integral


   $Version
5.1 for Microsoft Windows (October 25, 2004)

   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)]]

   Integrate[1/x^2, {x, 2, A}, Assumptions -> {A > 0}]
If[A > 2, 1/2 - 1/A, Integrate[1/x^2, {x, 2, A}, Assumptions -> A <= 2]]

   Integrate[1/x^2, {x, 2, A}, Assumptions -> {0 < A <= 2}]
1/2 - 1/A

You didn't say to which previous version of Mathematica you refer, but 
in 5.1, obviously Mathematica is being cautious -- and, I think (without 
my carefully checking all cases), correct!


richard speers 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.
> 
> 

-- 
Murray Eisenberg                     murray at math.umass.edu
Mathematics & Statistics Dept.
Lederle Graduate Research Tower      phone 413 549-1020 (H)
University of Massachusetts                413 545-2859 (W)
710 North Pleasant Street            fax   413 545-1801
Amherst, MA 01003-9305


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