Re: Integral confusion
- To: mathgroup at smc.vnet.net
- Subject: [mg107290] Re: [mg107262] Integral confusion
- From: Bob Hanlon <hanlonr at cox.net>
- Date: Mon, 8 Feb 2010 03:35:00 -0500 (EST)
- Reply-to: hanlonr at cox.net
f1[x_] = Integrate[1/(x + 1) - 1/(x + 6), x] // Simplify
log(-2 (x+1))-log(2 (x+6))
f1'[x]
1/(x + 1) - 1/(x + 6)
It means that for the result to be real that x must be less than -1.
Reduce[-2 (x + 1) > 0]
x < -1
Looking at both Logs
Reduce[{-2 (x + 1) > 0, 2 (x + 6) > 0}]
-6 < x < -1
Plot[f1[x], {x, -6, -1}]
If you don't like the Log form, use FullSimplify
f2[x_] = Integrate[1/(x + 1) - 1/(x + 6), x] // FullSimplify
-2*ArcTanh[(2*x)/5 + 7/5]
f2'[x] // Simplify // Apart
1/(x + 1) - 1/(x + 6)
f1[x] == f2[x] // FullSimplify
True
Bob Hanlon
---- Jon Joseph <josco.jon at gmail.com> wrote:
=============
All: Is this integral wrong? If not could someone explain the minus sign
inside the log?
Integrate[1/(x + 1) - 1/(x + 6), x] // Simplify
log(-2 (x + 1)) - log(2 (x + 6))
Thanks, Jon.=