Dear Jakearle,
Thanks for your question. An excellent resource for computing
indefinite integrals is the following Integrator from Wolfram
Reseaech:
http://integrals.wolfram.com/
The site fills in a random sample, which you can change to anything
you like. For infoormation on how to enter a formula, see these tips:
http://integrals.wolfram.com/about/input/
The input format is, not surprisingly, the same as that for
Mathematica, if you are familiar with this software.
The input is the following:
(1/x) / Log[a+bx]
Using this tool, the integral is
(Log x) / Log (a+bx)
(Log here is the natural log, aka ln.)
Another integral lookup tool can be found here, at Berkeley:
http://torte.cs.berkeley.edu:8010/tilu
I hope this was helpful.
Best,
weltega 
Clarification of Answer by
weltega
on
14 May 2005 07:27 PDT
My apologies... I misunderstood the statement of your question. The
input should be:
(Log[a+bx])/x
The resulting integral is then:
Log[a+bx] * Log[x]
weltega

Clarification of Answer by
weltega
on
14 May 2005 07:54 PDT
OK... Must not be my day. Don't try to do math before coffee!
After playing around with the Wolfram integration tool, I realized
that it doesn't recognize implicite multiplication without a space
between the variables. For this reason, in input I gave you, "bx" was
treated as a constant, rather than as b times x. This can be fixed by
either putting a space in (b x) or making the multiplication explicite
(b*x). The correct input should be:
(Log[a+b*x])/x
This results in a more complicated indefinite integral, as one might expect:
Log[x] (Log[a+b*x]  Log[1+b*x/a])  PolyLog[2, bx/a].
The PolyLog (Polylogarithm) function, also known as Jonquière's
function, is described in detail here:
http://mathworld.wolfram.com/Polylogarithm.html
and here:
http://functions.wolfram.com/ZetaFunctionsandPolylogarithms/PolyLog/
Using traditional notation, then, one can substitute
PolyLog[2, b x/a] = Li_2 [b x/a]
Also, Li_2 [b x/a] can be written as
Sum (from k=1 to infinity) [((b x/a)^k)/k^2]
Sorry for the confusion. Let me know if you require any clarification.
weltega

Request for Answer Clarification by
jakearlega
on
15 May 2005 16:03 PDT
Shouldn't that simplify to ln(a)*ln(x)  PolyLog[2, bx/a],
since ln(1 + bx/a) = ln[(a + bx)/a] = ln(a + bx)  ln(a)?

Clarification of Answer by
weltega
on
15 May 2005 19:56 PDT
That is correct.
weltega
