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Q: Calculus - Indefinite integral ( Answered,   0 Comments )
Subject: Calculus - Indefinite integral
Category: Reference, Education and News > Homework Help
Asked by: jakearle-ga
List Price: $2.00
Posted: 13 May 2005 10:31 PDT
Expires: 12 Jun 2005 10:31 PDT
Question ID: 521326
Indefinite integral of 1/x times natural log of (a + bx).
a and b are real numbers.
Subject: Re: Calculus - Indefinite integral
Answered By: welte-ga on 14 May 2005 07:18 PDT
Dear Jakearle,

Thanks for your question.  An excellent resource for computing
indefinite integrals is the following Integrator from Wolfram

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:

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:

I hope this was helpful.



Clarification of Answer by welte-ga on 14 May 2005 07:27 PDT
My apologies... I misunderstood the statement of your question.  The
input should be:


The resulting integral is then:

Log[a+bx] * Log[x]


Clarification of Answer by welte-ga 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:


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:

and here:

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.


Request for Answer Clarification by jakearle-ga 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 welte-ga on 15 May 2005 19:56 PDT
That is correct.

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