- #1

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But, what it would be, if [itex]x[/itex] is negative?

I calculated this way,

[itex]\int \frac {1}{x} \, dx = - \int \frac {1}{-x} \, dx = \int \frac {1}{-x} \, d(-x) = ln (-x) + C[/itex] [because, [itex]-x[/itex] has a positive value.]

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- Thread starter arpon
- Start date

- #1

- 236

- 16

But, what it would be, if [itex]x[/itex] is negative?

I calculated this way,

[itex]\int \frac {1}{x} \, dx = - \int \frac {1}{-x} \, dx = \int \frac {1}{-x} \, d(-x) = ln (-x) + C[/itex] [because, [itex]-x[/itex] has a positive value.]

- #2

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- #3

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- #4

pwsnafu

Science Advisor

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##ln(-x)## is a valid expression for negative x.

##\ln(ab) = \ln(a) + \ln(b)## is a property which is valid for positive a and b, it is not a definition of the logarithm. The property being unsatisfied for negatives has nothing to do with whether ##\ln(-x)## is defined or not.

Last edited:

- #5

Svein

Science Advisor

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To maximize the confusion: **ln(-x) = ln((-1)*x) = ln(-1) + ln(x) = iπ + ln(x)**

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