# How do you find the integral of (e^(-1/x)) / x^2?

Jun 20, 2015

At first, I was thinking why it was written like this. It makes sense... and this is NOT integration by parts. This can be done by u-substitution.

$= \int {e}^{\text{-1/x"/x^2dx = int e^"-1/x}} \cdot \frac{1}{x} ^ 2 \mathrm{dx}$

Let:
$u = {e}^{\text{-1/x}}$
$\mathrm{du} = {e}^{\text{-1/x}} \cdot \frac{1}{x} ^ 2 \mathrm{dx}$

This is rather interesting; there's no need to use $u$ here. We have the whole thing as $\mathrm{du}$.

$\implies \int \mathrm{du}$

$= u + C$

$= {e}^{\text{-1/x}} + C$