The integral $∫e^x\left(1+\frac{1}{x}+log\, x\right)$ is equal to :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $e^x(1+log \, x) + C$
$∫e^x+e^x\left(1+\frac{1}{x}+\log x\right)$
$=e^x+∫e^x\left(1+\frac{1}{x}+\log x\right)$
$⇒e^x+e^x\log x+C$
$=e^x(1+\log x)+C$