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)

```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.=

```

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