Solution :
We have, I = \(\int\) \(log {1\over x}\) dx
I = \(\int\) \(log 1 – log x\) dx = \(\int\) (-log x) dx
By using integration by parts formula,
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
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