∫ x^x≤ft(1+log _e x) d x is equal to

The correct answer is Option (3) → $x^x+C$

$I=\int x^x\left(1+\log x\right) d x$

and,

$x^x=e^{x\log x}$

$⇒I=\int e^{x\log x}(1+\log x)dx$   ...(1)

let,

$t=x\log x$

$⇒\frac{dt}{dx}=\log x+1$

Substituting this in (1),

$I=\int e^tdt$

$=e^t=e^{x\log x}+C$

$=x^x+C$