the intergral of: [e^(1/x)]/x^2 dx

\(\displaystyle \frac{e^{\frac{1}{x}}}{x^2} = e^{x^{-1}}x^{-2}\).

To find \(\displaystyle \int{e^{x^{-1}}x^{-2}\,dx}\)

Let \(\displaystyle u = x^{-1}\) so that \(\displaystyle \frac{du}{dx} = -x^{-2}\).

\(\displaystyle \int{e^{x^{-1}}x^{-2}\,dx} = -\int{e^{x^{-1}}(-x^{-2})\,dx}\)

\(\displaystyle = -\int{e^{u}\,\frac{du}{dx}\,dx}\)

\(\displaystyle = -\int{e^u\,du}\)

\(\displaystyle = -e^{u} + C\)

\(\displaystyle = -e^{x^{-1}} + C\)

\(\displaystyle = -e^{\frac{1}{x}} + C\).