Question

$\int e^{-{\log }_{e}x}dx$ is

Answer 1

Now,

$\int e^{-{\log }_{e}x}dx$

$=$$\int e^{{\log }_e\left(\frac{1}{x}\right)}dx$

$=\int \frac{1}{x}dx$

$=\log |x|+c$ [ $c$ being integrating constant]