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

[Jonathan King <EMAIL:PROTECTED>]

On Thu, 1 Nov 2001, 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. :-)

[snip]

Well actually, http://integrator.com (WOW! I'm in LOVE.) came to our
rescue in this case.  You have to enter your input in Mathematica format,
but then it just works.  Obviously, the answer we were seeking was:

    \frac{1}{(a-1)(1+x)}
       - \frac{a \log(1+x)}{(a-1)^2}
       + \frac{a \log(1+ax)}(a-1)^2}

Sorry for the oversight; hope this helps somebody else.

jking


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