Integral of 1/x: Negative x Solution

[arpon]

As we know, \int \frac {1}{x} \, dx = ln (x) + C
But, what it would be, if x is negative?
I calculated this way,
\int \frac {1}{x} \, dx = - \int \frac {1}{-x} \, dx = \int \frac {1}{-x} \, d(-x) = ln (-x) + C [because, -x has a positive value.]

 

[Joshua L]

The variable $x$ can be either positive or negative because the fact is, \int \frac{1}{x} dx = \ln({|x|}) + C

 

[dextercioby]

The function 1/x is (maximally) defined at all points in R minus 0, the natural logarithm only at points >0. It turns out that $\ln -x$ is not a valid expression in the reals, because it conflicts with the known $\ln ab = \ln a + \ln b$ by being forced to consider $\ln -1$.

 

[pwsnafu]

The function 1/x is (maximally) defined at all points in R minus 0, the natural logarithm only at points >0. It turns out that $\ln -x$ is not a valid expression in the reals, because it conflicts with the known $\ln ab = \ln a + \ln b$ by being forced to consider $\ln -1$.

$ln(-x)$ 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.

 

[Svein]

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