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i hope i make this clear and to the point. there is an integral that is bothering me. i will express it as the combination of 2 integration formulas that you can look up, under logarithmic functions (http://en.wikipedia.org/wiki/List_of_integrals_of_logarithmic_functions) [Broken]. here are two identities from that list:

[tex]\int \ln(cx) dx = x\ln(cx) - x[/tex]

and

[tex]\int \ln(ax + b) dx = x\ln(ax + b) - x + (b/a)\ln(ax + b) [/tex]

I have an equation(it actually comes from a famous paper in physics) that basically represents the difference between the two. To make things simpler, a = c = 1 in my equation.

So I have:

(1) [tex]\int \ln(x + b) dx - \int \ln(x) dx[/tex]

using the identities I have:

[tex]x\ln(x + b) - x + (b)\ln(x + b) - x\ln(x) + x[/tex]

[tex]x\ln(x + b) + (b)\ln(x + b) - x\ln(x)[/tex]

[tex](x + b)\ln(x + b) - x\ln(x)[/tex]

so that's that. Now let me do it slightly differently(and don't ask why):

I will factor out the 'b' from the equation first (1):

(2) [tex]\int \ln[(x/b + 1)*(b)] dx - \int \ln[(x/b)*b] dx[/tex]

[tex]\int \ln[(x/b + 1) + ln(b)] dx - \int \ln[(x/b) + ln(b)] dx[/tex]

[tex]\int \ln(x/b + 1) dx - \int \ln(x/b)dx + \int ln(b)dx - \int ln(b)dx] dx[/tex]

[tex]\int \ln(x/b + 1) dx - \int \ln(x/b) dx[/tex]

using the same identities

[tex]x\ln(x/b + 1) - x + b\ln(x/b + 1) - x\ln(x/b) + x[/tex]

[tex]x\ln(x/b + 1) + b\ln(x/b + 1) - x\ln(x/b)[/tex]

[tex](x + b)\ln(x/b + 1) - x\ln(x/b)[/tex]

this is now a different result from approach (1). You can't, unless I'm wrong, recover it again by resubstituting the factor b in again.

My question: why does this simple factoring out, change the answer here?

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# Two different results for the same integral

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