Question

The value of $\int e^{-\ln x^2}dx$ is

Answer 1

$\int e^{-\ln x^2}dx$

$=\int e^{\ln x^{-2}}dx$

$=\int x^{-2}dx$

$=\frac{x^{-2+1}}{-2+1}+c$... where $c$ is the constant of integration

$=\frac{x^{-1}}{-1}+c$

$=\frac{-1}{x}+c$