The value of the integral ∫e^x(log x+1/x)dx is:

The correct answer is Option (1) → $e^x\log\, x+C$, Where C is a constant.

$∫\left(log\, x+\frac{1}{x}\right)^{e^x}dx$

$=e^x\log x+C$ as $f(x)=\log x$

$f'(x)=\frac{1}{x}$

so $\int e^x(f(x)+f'(x))dx$

$=e^xf(x)+C$