# 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

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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