The integral ∫logxdx is equal to :

The correct answer is Option (3) → $x\log x -  x + c$

$I=\int\log xdx$

$=x\log x-\int\left(\frac{d}{dx}(\log x)\right)x\,dx$

$=x\log x-\int\frac{x}{x}dx$

$=x\log x-x+c$