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.

\displaystyle \int \frac{1}{x}\,dx=\ln(|x|)+C_0=\ln(|x|)+\ln(2)+C_1\,,\ 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 ln(x) is undefined. That it applies can be seen by forming the derivative in the separate cases when x > 0 and x < 0 respectively. One should still watch for intervals which include zero for there neither ln(x) nor ln|x| are defined.

 

[phospho]

thanks!