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 ```Indefinite integral of 1/x times natural log of (a + bx). a and b are real numbers.```
 ```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, -welte-ga``` 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: (Log[a+bx])/x The resulting integral is then: Log[a+bx] * Log[x] -welte-ga``` 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: (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. -welte-ga``` 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. -welte-ga```