Answer
$\dfrac{1}{2}e^{-\frac{1}{x^2}} +c$
Work Step by Step
Solve $\int \dfrac{e^{-1/x^2} }{x^3}dx$
Let us $p=-\dfrac{1}{x^2}$ and $dp=2(\dfrac{1}{x^2})dx$
This implies $\int \dfrac{e^{-1/x^2} }{x^3}dx=\dfrac{1}{2}\int e^p dp$
$=\dfrac{1}{2} e^p +c$
$=\dfrac{1}{2}e^{-\frac{1}{x^2}} +c$
Hence,$\int \dfrac{e^{-1/x^2} }{x^3}dx=\dfrac{1}{2}e^{-\frac{1}{x^2}} +c$
