What is the integration of #1/x#?

1 Answer
Apr 11, 2015

#int 1/x dx = ln abs x +C#

The reason depends on which definition of #ln x# you have used.

I prefer:
Definition: #lnx = int_1^x 1/t dt# for #x>0#

By the Fundamental Theorem of Calculus, we get: #d/(dx)(lnx) = 1/x# for #x>0#

From that and the chain rule, we also get #d/(dx)(ln(-x)) = 1/x# for #x<0#

On an interval that excludes #0#, the antiderivative of #1/x# is
#lnx# if the interval consists of positive numbers and it is #ln(-x)# if the interval consists of negative numbers.

#ln abs x# covers both cases.