Intergrals and e

**calcutron20** · May 9th 2010, 07:01 PM

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

**Prove It** · May 9th 2010, 07:06 PM

Originally Posted by calcutron20

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

$\frac{e^{\frac{1}{x}}}{x^2} = e^{x^{-1}}x^{-2}$ .

To find $\int{e^{x^{-1}}x^{-2}\,dx}$

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

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

$= -\int{e^{u}\,\frac{du}{dx}\,dx}$

$= -\int{e^u\,du}$

$= -e^{u} + C$

$= -e^{x^{-1}} + C$

$= -e^{\frac{1}{x}} + C$ .

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