What is the integration of log 1/x ?

Solution :

We have, I = \(\int\) \(log {1\over x}\) dx

I = \(\int\) \(log 1 – log x\) dx = \(\int\) (-log x) dx

Let I = -(\(\int\) log x .1) dx

where log x is the first function and 1 is the second function according to ilate rule.

I = – (log x . {\(\int\) 1 dx} – \(\int\) { \(d\over dx\) (log x) . \(\int\) 1 dx }) dx

I = – {(log x) x – \(\int\) \(1\over x\).x }dx

= – x (log x) + \(\int\) 1 dx

= – x (log x) + x + C = x – x log x + C

Hence, the integration of log 1/x with respect to x is x – log x + C