Re: [MLUG - DISCUSSION] Yet Another Wacky Integral...

[Jonathan King <EMAIL:PROTECTED>]

On Thu, 1 Nov 2001, Stephen Montgomery-Smith wrote:

> Jonathan King wrote:
> >
> > OK, so we need to solve the following integral:
> >
> >    \int_0^z\frac{dx}{(1+x)^2(1+ax)}
> >
> > No money involved; just your Geek Pride. :-)
>
> Partial fractions are your friend - in Mathematica the command
> Apart[1/((1 + x)^2(1 + a x))]
>
> 1/((1 - a)*(1 + x)^2)
>   - a/((-1 + a)^2*(1 + x))
>   + a^2/((-1 + a)^2*(1 + a*x))
>
> (but it would be quite easy to do by hand).

Yes, I did feel really silly when I saw the answer at integral.com
and instantly realized this was much easier than I'd thought...usually
when Jeff has an integral to do, it turns out way more obnoxious, and
he was pretty confident about this one being nasty.

> So the indefinite integral evaluates to:
>
> 1/((-1 + a)*(1 + x)) - (a*Log[1 + x])/(-1 + a)^2 +
>   (a*Log[1 + a*x])/(-1 + a)^2
>
> Mathematica and Maple are both available for Linux, and MU has a site
> license, so faculty can use it for about $25 per computer per year.

Well, now *that's* cool news I can use. :-)  Meanwhile, did anybody ever
find a central "site licenses at MU page" we could all browse?

> > pps--Thanks ahead of time to Stephen Montgomery-Smith. :-)
> >      --Jeff Rouder and Jon King
>
> Best to also email me directly because I don't always follow mlug.

Me neither...plus, I thought maybe somebody else could suggest an on-line
resource.  As it turns out, http://integral.com was all that was needed
here...

jking


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