TuringTest
  • TuringTest
we can see that\[\frac d{dx}\ln(ax)=\frac1x\]so why is\[\int\frac{dx}{x}\neq\ln(ax)+C\]why can we omit the possible constant, a?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Callisto
  • Callisto
i think you've omitted the constant a in derivative
campbell_st
  • campbell_st
they cancel each other... eg... y = ln(3x) dy/dx = 3/3x = 1/x
TuringTest
  • TuringTest
yeah^ thank you

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TuringTest
  • TuringTest
every friggin' time they cancel, so why do we exclude it from the integral?
experimentX
  • experimentX
i think you are adding two constants on second integration 'a' and 'C' I think you can remove 'a' and write 'ln a' inplace of C, you will get the same result
TuringTest
  • TuringTest
but\[\ln|ax|+C\neq\ln|x|+C\]if \(a\neq0\)
TuringTest
  • TuringTest
I mean if \(a\neq1\)
mysesshou
  • mysesshou
my book has \[\frac{d}{dx}(ln (u)) = \frac{u'}{u}\] ... so I guess if you use that one, then you would still have the a constant, when u=ax Also, I have \[\frac{du}{u} = ln \left| u \right| +C\] which would also show the a/ax where it also cancels. where u=ax But, yeah, that is interesting.
experimentX
  • experimentX
∫dxx=ln(x)+C , since 'C' is a constant, you can equivalently write 'ln(a)' for it, and you get ln(ax) which is again the general form
TuringTest
  • TuringTest
now that makes sense! thank you @experimentX
experimentX
  • experimentX
you are welcome
mysesshou
  • mysesshou
That does make sense for C=ln(a) ... Never really thought about the why's lately :)

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