Derivative and integral of the natural log

[phospho]

not really a problem, but more curious

if we differentiate ln(2x) we get 2/(2x) = 1/x by the chain rule, but if we integrate 1/x we get ln|x|? Could anyone explain why this is the case, thanks.

 

[SammyS]

not really a problem, but more curious

if we differentiate ln(2x) we get 2/(2x) = 1/x by the chain rule, but if we integrate 1/x we get ln|x|? Could anyone explain why this is the case, thanks.

ln(2x) = ln(x) + ln(2)

Don't forget the constant of integration.

[itex]\displaystyle \int \frac{1}{x}\,dx=\ln(|x|)+C_0=\ln(|x|)+\ln(2)+C_1\,,\ [/itex] where C₀ = C₁ + ln(2) .

 

[Ferramentarius]

As to the absolute value it is a generalization which works in the case of negative numbers for which the real logarithm [itex] ln(x) [/itex] is undefined. That it applies can be seen by forming the derivative in the separate cases when [itex] x > 0 [/itex] and [itex] x < 0 [/itex] respectively. One should still watch for intervals which include zero for there neither [itex] ln(x) [/itex] nor [itex] ln|x| [/itex] are defined.

 

[phospho]

thanks!