The value of ∫limits_1^e log x dx is:

The correct answer is Option (2) → 1

$I = \int\limits_{1}^{e} \log x \, dx$

$= [x \log x - x]_1^e$

$= [e \log e - e] - [1 \log 1 - 1]$

$=e\log e-e-\log 1+1$ {Since $\log e = 1, \log 1 = 0$}

$= e - e - 0 + 1$ 

$I = 1$