Question

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

Answer 1

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^"-1/x"/x^2dx = int e^"-1/x"*1/x^2dx`

Let:
`u = e^"-1/x"`
`du = e^"-1/x"*1/x^2dx`

This is rather interesting; there's no need to use `u` here. We have the whole thing as `du`.

`=> int du`

`= u + C`

`= e^"-1/x" + C`