       Re: Integral confusion

• To: mathgroup at smc.vnet.net
• Subject: [mg107299] Re: [mg107262] Integral confusion
• From: Jon Joseph <josco.jon at gmail.com>
• Date: Mon, 8 Feb 2010 03:36:38 -0500 (EST)
• References: <20100207083821.M5V2P.501735.imail@eastrmwml34>

```Thanks Bob and David. I completely understand your responses and I
figured that was what Mathematica was doing. However, if I asked my
students to integrate 1/(x+1) - 1/(x+6) and they came back with the
answer Mathematica found I would, at the very least, call that student
over for a discussion. Log arguments in absolute value are certainly

I appreciate Mathematica's complexity but it is unfortunate that a user
has to piece together the underlying logic on such a simple problem. Jon

On Feb 7, 2010, at 7:38 AM, Bob Hanlon wrote:

>
> 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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